# Talk:Continuum hypothesis

## who believes CH or ~CH ?

I don't entirely agree with the following passage:

Generally speaking, mathematicians who favor a "rich" and "large" universe of sets are against CH, while those favoring a "neat" and "controllable" universe favor CH.

See the Maddy reference I added, page 500, the sections entitled Not-CH is restrictive (in favor) and Modern forcing (in favor).

I'll try to come up with some wording that's not too awkward, illustrating the historical view reported in the existing Wiki article, while also pointing out that between models having all the same reals, it's the ones with more sets of reals that are more likely to satisfy CH. --trovatore

My attempt is now in place Trovatore 05:04, 27 Jun 2005 (UTC)

User Martindowd has been arguing that the truth of CH, indeed V=L, should be taken more seriously. He suggests adding the following at the end of the second paragraph of the "Arguments for and against CH" section of the Continuum hypothesis page. If you have any opinion on this, please let me know.

Recently, arguments have appeared that this view should be re-evaluated, and that there are arguments in favor of both (Dowd 2011).

Dowd, Martin (2011) "Some New Axioms for Set Theory", http://www.ijpam.eu/contents/2011-66-2/1/1.pdf Martindowd (talk) 15:04, 21 September 2011 (UTC)

See Constructible universe#L and large cardinals. JRSpriggs (talk) 17:34, 21 September 2011 (UTC)
This is a related topic; but the paper in question takes a new approach.Martindowd (talk) 17:52, 24 September 2011 (UTC)

### Did Cantor believe CH?

I would add to Trovatore's point that it is questionable (as stated on the page) that Cantor himself believed CH. Did he not alternate back and forth during his manic (and depressive) phases, sometimes believing CH, sometimes believing ~CH? See the BBC TV documentary "Dangerous Knowledge" willbown. 16 June 2008. —Preceding comment was added at 22:37, 16 June 2008 (UTC)

I'd never heard that he ever believed ¬CH, only that he sometimes despaired of ever being able to prove CH. Interesting if true, though. What did the documentary say exactly? Has anyone seen this claim in any source available online or at the library? --Trovatore (talk) 22:53, 16 June 2008 (UTC)
I've got the programme on tape, and I'll look it up and report back. Great programme, BTW. --Michael C. Price talk 18:08, 24 August 2008 (UTC)
The programme hints that Cantor wanted to believe in CH -- but it is not explicit. As for Cantor's alternating between CH and ¬CH, that was in reference to him thinking that he found a proof of CH at times and ¬CH at other times. Proof and belief are not quite the same.--Michael C. Price talk 08:25, 25 August 2008 (UTC)
In 1978 Cantor wrote in "Ein Beitrag zur Mannigfaltigkeitslehre" (A contribution to the theory of manifolds)
Durch ein Inductionsverfahren, auf dessen Darstellung wir hier nicht näher eingehen, wird der Satz nahe gebracht, daß die Anzahl der nach diesem Eintheilungsprinzip sich ergebenden Klassen eine endliche, und zwar, daß sie gleich zwei ist [...] Eine genaue Untersuchung dieser Frage verschieben wir auf eine spätere Gelegenheit.
Free translation: An induction process/method, which we will not present here in detail, suggests the theorem that the number of classes which result from the classification principle [i.e., equivalence classes modulo bijections, of infinite subsets of the real line] is finite, and in fact 2. [...] We will resume this investigation at a later moment.
I think this makes it clear that he thought he could work out a proof of CH. --Aleph4 (talk) 19:42, 9 May 2011 (UTC)
Little-known fact: When he wrote the above, he was blasting Stayin' Alive from his boombox. Could have interfered with his concentration. --Trovatore (talk) 20:24, 9 May 2011 (UTC)
Indeed, Stayin' Alive, top of the charts in 1978. I meant 1878 of course. --Aleph4 (talk) 15:17, 30 July 2011 (UTC)

cantor did not i can prove it.welcome to the university of night vale. [vince sippola] 99.249.188.5 (talk) 17:55, 28 December 2013 (UTC) don't believe me?i am vince, welcome to night vale, google me, i am vince Dragonbron, vince sippola, fgh, fgh34@hotmail.com, call me;)*99.249.188.5 (talk) 01:10, 29 December 2013 (UTC)

## Investigating the continuum hypothesis

I've removed this section:

If a set S were found that disproved the continuum hypothesis, it would be impossible to make a one-to-one correspondence between S and the set of integers, because there would always be elements of set S that were "left over". Similarly, it would be impossible to make a one-to-one correspondence between S and the set of real numbers, because there would always be real numbers that were "left over".

It's a good thing to add some intuition, but I don't know that this passage helped much. CH isn't about whether such a set S can be found in any ordinary sense, but just about whether one exists. --Trovatore 21:12, 22 October 2005 (UTC)

Since CH cannot be disproven, a set that denies it cannot be found. This doesn't mean it does not exist, but we're close :-) Honnza (talk) 06:01, 17 May 2008 (UTC)

CH cannot be disproved from ZFC. No, I'm afraid that's not particularly close to settling the issue. --Trovatore (talk) 08:40, 17 May 2008 (UTC)

## Any set to demonstrate CH?

The undecidability of CH begs the question: If there is a cardinality between ${\displaystyle \aleph _{0}}$ and ${\displaystyle 2^{\aleph _{0}}}$, then what sets might there be that have this cardinality?

In other words, is there any known set that is larger than ${\displaystyle \aleph _{0}}$, but such that you need to set the truth value of CH to determine whether or not it's equivalent to continuum?

Moreover, is there any evidence pointing to the existence or non-existence of such a set, or whether it would be possible to find it if it does exist? -- Smjg (talk) 16:35, 28 February 2008 (UTC)

Well, you have to keep in mind here the distinction between a set and a definition of a set. Think of sets as collections of things just lumped together at random, not necessarily in accordance with any rule. Then it might happen, just by accident, that there's some rule such that everything in the set satisfies the rule, and everything outside the set does not -- in that case we say that the set is "definable", but you still can't identify the set with its definition; that way lies all sorts of trouble (ask Frege).
So rephrasing your question -- is there a definition for a set of reals such that ZFC neither proves nor disproves that the set of all reals satisfying the definition, has cardinality ${\displaystyle 2^{\aleph _{0}}}$ (but does prove that the set is uncountable)? Sure, but possibly not a terriby interesting definition. Something like "the first ${\displaystyle \aleph _{2}}$ ordinal-definable reals in the natural wellorder on OD, or all the reals if there are not ${\displaystyle \aleph _{2}}$ OD reals". --Trovatore (talk) 18:07, 28 February 2008 (UTC)
Also, ${\displaystyle L_{\omega _{1}}\cap {\mathcal {P}}(\omega )}$ is the set of all subsets of the natural numbers (sometimes identified with the reals) which are constructed (see Constructible universe) before the first uncountable stage. It has cardinality ${\displaystyle \aleph _{1}}$, I believe. JRSpriggs (talk) 08:34, 29 February 2008 (UTC)
No, not unless ${\displaystyle \aleph _{1}^{L}=\aleph _{1}}$ (which it doesn't, of course). --Trovatore (talk) 18:42, 29 February 2008 (UTC)
To Trovatore: Thanks for the correction. JRSpriggs (talk) 12:03, 1 March 2008 (UTC)
The easiest example of a set that has cardinality ${\displaystyle \aleph _{1}}$ is the set of all countable ordinals (where everything is relativized to a fixed model of ZFC). Each countable ordinal can be viewed as a linear order on the natural numbers and thus an element of ${\displaystyle 2^{\omega }}$. So if you use AC to choose for each countable ordinal a single real encoding that ordinal, you will obtain a set that has cardinality ${\displaystyle \aleph _{1}}$, but will not have the cardinality of the continuum unless CH holds. — Carl (CBM · talk) 15:03, 1 March 2008 (UTC)
Quoting Carl (an inch above) "Each countable ordinal can be viewed as a linear order on the natural numbers..." Is the converse statement "Each linear order on the natural numbers can be viewed as a countable ordinal" necesarilly true? If yes, is there then a one-to-one correspondence between linear orders on the natural numbers and countable ordinals? Or is that last statement perhaps equivalent to the CH? (When thinking of the Cantor-Bernstein theorem, it might be that way. (I'm just speculating now of course.)) The number of orderings of a finite number n is n!. Can one by a naive extension of the factorial function say that CH can be "rephrased" as the statement ${\displaystyle \omega }$! = ${\displaystyle \omega _{1}}$? YohanN7 (talk) 17:40, 18 May 2008 (UTC)
Taking the last question first: Yes, that's a reasonable way of putting it, though not quite by the reasoning you used; the reasoning has some flaws but it's close enough to see that the factorial of ${\displaystyle \aleph _{0}}$ (not a term that's used much but it's pretty clear what it means) is ${\displaystyle 2^{\aleph _{0}}}$.
Getting more into the details, more specifically, not every linear order of the natural numbers corresponds to a countable ordinal, but only the wellorderings. To a wellordering of the naturals you associate a unique countable ordinal, namely the length of the wellordering. And going the other direction, any (infinite) countable ordinal can be represented as the length of some wellordering--but not a unique one; there will be lots of different wellorderings with the same length. So what this gets us is that there are at least as many wellorderings of the naturals as there are countable ordinals (strictly speaking, infinite countable ordinals, but the finite ones are easily accounted for). It doesn't tell us there aren't more. --Trovatore (talk) 18:20, 18 May 2008 (UTC)
Interesting! To quote yourself Trovatore, "It's a good thing to add some intuition...", but you are always very (too?;) careful to make intuitive arguments close to precise mathematical statements. I believe that the above reply does shed additional light on what the CH actually is and goes beyond intuition. (It actually sounds like a statement of ZF(C) in my ears, making it mathematics.) I also think that som revised form of it may qualify for the main article as an example of what CH and its negation would imply. Most people who knows what a one-to-one correspondence will know what a wellordering is.YohanN7 (talk) 22:34, 20 May 2008 (UTC)

## Independence of CH from large cardinal axioms

So far, CH appears to be independent of all known large cardinal axioms in the context of ZFC.

But in the context of ZF, j:V into V refutes choice (according to the article on Reinhardt cardinals anyway), and GCH implies it (according to this article). So then j:V into V would refute GCH. Right? --Unzerlegbarkeit (talk) 15:58, 27 May 2008 (UTC)

It may be that J:V into V refutes ZF.Kope (talk) 16:29, 27 May 2008 (UTC)
Well, sure. It may be that any large cardinal axiom refutes ZF. But so far no such refutation is known. Nor is any proof, refutation, or independence of CH or GCH from any large cardinal axiom known, other than this one. Is that correct? --Unzerlegbarkeit (talk) 02:36, 28 May 2008 (UTC)
The Generalized Continuum Hypothesis is much stronger than the Continuum Hypothesis. So many propositions may be consistent with CH, but not with GCH. Reinhardt's cardinal cannot exist in ZFC so the sentence in question does not apply to it. JRSpriggs (talk) 06:37, 28 May 2008 (UTC)
Sure. And I did say "in the context of ZF" and "GCH". I guess my implicit assertion is that whatever makes the sentence in question worth mentioning makes this worth mentioning too, assuming I've stated the situation correctly. --Unzerlegbarkeit (talk) 10:24, 28 May 2008 (UTC)
(To Unzerlegbarkeit ) Not quite. A favorable solution would be to show that the consistency of some (axiom of choice) large cardinal implies the consistency of j:V into V, minus AC. Incidentally, GCH has more than one formulations, which are equivalent in the presence of AC, but not necessarily so in the absence of it. One of them implies AC. Therefore, I think, it is more correct to say that one form of GCH implies AC. Kope (talk) 06:45, 28 May 2008 (UTC)
To Kope: See Talk:Axiom of choice#GCH implies AC?? for a proof that GCH implies AC. This establishes that there is only one version of GCH, even in ZF (without choice being assumed otherwise). JRSpriggs (talk) 06:59, 28 May 2008 (UTC)
But CH doesn't, so might it have inequivalent formulations in the absence of AC? Is this also worth mentioning? --Unzerlegbarkeit (talk) 10:24, 28 May 2008 (UTC)
Yep. The following are not equivalent in the absence of AC:
The first two are considered somewhat standard formulations of CH; I only included the latter two so that it's clear they are all the same under the axiom of choice. — Arthur Rubin (talk) 17:22, 28 May 2008 (UTC)
Hmmm. In ZFU (ZF with Urelements), AH (aka Aleph Hypothesis) ${\displaystyle \left(2^{\aleph _{\alpha }}=\aleph _{\alpha +1}\right)}$ does not imply GCH (if X is an infinite set, there is no Y such that ${\displaystyle X (with "<" in the sense of cardinality)). (Source: my mother's book Set Theory for the Mathematician, I believe). I'm sure that fact that PW (the power set of a well-ordered set can be well-ordered, a trivial consequence of AH) does not imply AC in ZFU, although it does in ZF, is in both editions of my parents' book Equivalents of the Axiom of Choice. — Arthur Rubin (talk) 17:40, 28 May 2008 (UTC)
However, see Beth number#Generalization — if the ur-elements form a set which is equinumerous with a pure set (a set whose transitive closure contains no ur-elements), then I believe that ZFU+AH would imply AC and thus GCH. JRSpriggs (talk) 14:08, 29 May 2008 (UTC)
Yes, I believe that's correct. The proof escapes me at the moment.... — Arthur Rubin (talk) 14:34, 29 May 2008 (UTC)
Consider the class of pure sets. It satifies ZF+AH. Since bijections between pure sets are themselves pure sets, it has the same aleph numbers as the original model (which included urelements). So the class of pure sets satifies AC. Thus the pure set which is equinumerous with the set of urelements can be well ordered. Thus the set of urelements can be well ordered. Then do induction on the rank α to show that Vα can be well ordered. So the universe satisfies AC. So it satisfies GCH. OK? JRSpriggs (talk) 15:31, 29 May 2008 (UTC)

### Forms of CH without AC

The form ${\displaystyle \neg \exists m\;\aleph _{0} makes the continuum hypothesis meaningful even if the continuum isn't well-ordered. The article states:

The continuum hypothesis is closely related to many statements in analysis, point set topology and measure theory. As a result of its independence, many substantial conjectures in those fields have subsequently been shown to be independent as well.

Do we have any examples which would make sense without AC, and would they depend on the form of CH? Also, on the other side of the coin, my understanding is that there could be no purely arithmetic consequences, roughly because ZF + V=L proves CH anyway (and even GCH), and relativising everything to L makes no difference to (first-order) arithmetic. If correct, I believe this is worth stating. Also, the article states:

Assuming the axiom of choice, there is a smallest cardinal number ${\displaystyle \aleph _{1}}$ greater than ${\displaystyle \aleph _{0}}$, and the continuum hypothesis is in turn equivalent to the equality ${\displaystyle 2^{\aleph _{0}}=\aleph _{1}.}$

The "assuming the axiom of choice" bit should really attach to the "in turn equivalent to", because even from the axioms ZF without AC, aleph_1 exists and is an immediate successor of aleph_0, right? --Unzerlegbarkeit (talk) 02:39, 1 June 2008 (UTC)

Without AC, there could be other cardinals which are larger than ${\displaystyle \aleph _{0}}$ but incomparable with ${\displaystyle \aleph _{1}}$, for example, the cardinality of a set which is the union of natural numbers with a Dedekind finite infinite set. In this case, neither the new cardinal nor ${\displaystyle \aleph _{1}}$ can be said to be the smallest cardinal greater than ${\displaystyle \aleph _{0}}$. JRSpriggs (talk) 06:38, 1 June 2008 (UTC)

## Original Sources

I notice there aren't any references to Cantor's original writings. I'll try to dig some up, but has anyone seen any English translated letters, etc? Libertyblues (talk) 00:03, 3 August 2008 (UTC)

## Hilbert's first has nothing to do with ZFC per se

This statement is revealing a transfinite-platonic bias. without a formalization like ZFC, it is not clear that it is possible to give a meaning to the statement that the continuum even has an ordinal size. Before Cantor insisted it was true, an infinite collection like the set of real numbers was not considered to have a definite size, let alone a definite ordinal size, and indeed, after Cohen, it is again a widespread point of view that the continuum does not have a definite size as an ordinal.

The idea that the set of all real numbers has definite properties in a platonic realm can be classified as a type of "transfinite platonism", which is just what platonism usually means nowadays. But transfinite Platonism can be logically separated from "computational platonism" (I made that up, but it needs a name)--- the position that all computer programs either halt or do not halt. The second position is what people mean by platonism in practice--- that the results of computations with symbols have a meaning, and questions about their outcome have a truth value. it was the position of Paul Cohen, shared by most mathematicians, that all questions about the integers/computer-programs are decided by a strong enough axiom system, which is a way of saying that they have a truth value in an axiom-system independent way.

But you can believe this, and still be a formalist regarding transfinite set theory. The position might be called "formalist", but "formalist" conflates two notions: "formalist regarding uncountable ordinals and sets the size of the continuum" and "formalist regarding countable infinity". A "formalist regarding countable infinity" would, for example, consider a nonstandard models of Peano arithmetic to be just as "true" a model of the integers as the standard model. Such a formalist would believe that some statements about diophantine equations like Fermat's Last Theorem, are undecidable in an absolute sense. This is the type of straw-man formalist that people argue against.

A "formalist regarding uncountable ordinals" on the other hand, would say that the truth value of the 3N+1 conjecture is well defined, but statements like the continuum hypothesis have no truth value except relative to an axiom system. This is the type of formalism which the forcing models foist upon you. If you accept this philosophy, which many people do, then you would not regard the continuum hypothesis as independent of ZFC.

This is the philosophical position which Cohen alludes to in his book, hopefully stated clearly enough there so that this exposition will be recognized as a clarification. I have been trying to figure out how to say it clearly for a while.Likebox (talk) 17:22, 24 August 2008 (UTC)

Hilbert wanted a proof of the continuum hypothesis (he doesn't seem to have explicitly countenanced a refutation). A proof from what axioms, I'm afraid he didn't really say. But whatever they were, they certainly weren't the axioms of ZFC, because his speech was in 1900, and ZFC wasn't formulated until a couple of decades later.
As for the situation today, it is certainly true that there are folks who are realists about the naturals but formalists beyond that, and that they mostly will not consider CH to have a well-defined truth value outside of its provability or refutability in some formal theory. There is still, however, no clear reason that that formal theory ought to be ZFC. --Trovatore (talk) 01:06, 25 August 2008 (UTC)
I see your point, thanks for clarifying. But ZFC is not arbitrary--- it's sort of natural--- you have the function axiom, and everything else just builds up big ordinals in sequence. It's like Peano arithmetic, anything else is going to be roughly similar. Hilbert had formalized logic in mind, and probably knew approximately what a formalized set theory would look like, and I think he meant ZFC. But I defer to someone who actually read more Hilbert.Likebox (talk) 02:51, 25 August 2008 (UTC)
In one sense, you're right, ZFC is not arbitrary — all its axioms are well-motivated (in retrospect) by the picture of the cumulative hierarchy. What's arbitrary is cutting off precisely at the strength of ZFC, no less and no more.
Now, the most natural and well-accepted enhancements to ZFC, the large cardinal axioms, don't settle CH either, or at least the ones we know so far don't settle it. It doesn't really follow that we won't ever find ones that do (though they would have to be of a somewhat different character from the currently known ones).
On the other hand, there are extensions to ZFC that someone or other has considered natural at some point that do settle CH. For example the axiom of constructibility settles it positively, whereas the proper forcing axiom settles it negatively. The big difference is, I'm afraid, a Platonistic one: there's not as compelling a reason to believe that either of these is true (and indeed there are extremely good reasons to believe that the axiom of constructibility is false).
The most interesting current line of inquiry into the question is one started by Woodin, who has a subtle and difficult argument that, if accepted, would imply that CH is false. --Trovatore (talk) 03:49, 25 August 2008 (UTC)
You like repeating the party line! But of course, you know that I think it's rubbishy. And Woodin's approach is neither subtle or difficult. It's just determinacy.Likebox (talk) 06:30, 25 August 2008 (UTC)
If you truly understand Woodin's approach, I'd be happy if you'd teach it to me; I know only some vague generalities about it. But it's certainly not "just determinacy". In fact I don't see that it has that much to do with determinacy, directly. What he claims is that he has something analogous to determinacy, one level up; that just as projective determinacy "settles all natural questions" (whatever that means exactly—note that the quotes are not intended to attribute the phrase to Woodin) about ${\displaystyle H(\aleph _{1})}$, his approach similarly "should" settle all natural questions about ${\displaystyle H(\aleph _{2})}$ (one of which is CH).Now, just by the way, "just" determinacy can get extremely difficult and subtle. But Ω-logic is still quite another thing. --Trovatore (talk) 06:44, 25 August 2008 (UTC)
I thought he was just repeating determinacy, I guess I'm totally wrong. I didn't read his paper very closely at all.Likebox (talk) 10:31, 25 August 2008 (UTC)
I know (from a very superficial look at) 'The Higher Infinite' that there were some (at the time of their discovery) unexpected connections between large cardinal axioms and various versions of the axiom of determinacy, for what that's worth. Zero sharp (talk) 14:49, 25 August 2008 (UTC)

(deindent) I am pretty ignorant about determinacy--- I only have only the most superficial heresay knowledge, and no clear understanding. But I have a somewhat negative opinion anyway, probably unjustified.

Well, then just a friendly suggestion: refrain from making categorical statements that Woodin's approach is 'just determinacy' and from accusing people of 'just repeating the party line'. It just makes you look ignorant. —Preceding unsigned comment added by 67.118.103.210 (talk) 23:32, 26 August 2008 (UTC)
I am ignorant. So is everybody else. That's life.Likebox (talk) 01:53, 27 August 2008 (UTC)

When I think about the continuum hypothesis, it is linked in my mind to the notion of measure and probability--- the reason that many people have intuitions about the cardinality of the real numbers is that it is consistent to talk about randomly choosing real numbers in a way that is different from randomly choosing integers or elements of a well ordered set. You can't pick an integer uniformly at random--- the concept doesn't make sense. You need a probability distribution, and certain integers are more likely than others. The same goes for any countable ordinal--- you need to weight different positions with different probability.

But for the interval [0,1], the notion of picking a number at random seems to be perfectly well defined, because you can roll dice to find each digit in sucession. This converges to a real number, and any property of that number that has probability zero is false. If a random number can be thought of as an element of the mathematical universe, with the property of belonging or not belonging to any previously specified set, it means that this previously specified set has a well defined Lebesgue measure. So not only is the continuum hypothesis false with random reals, it is meaningless, because the real numbers can't be well ordered. Any ordinal, no matter how large, can be imbedded in the reals by inductively mapping each successive element into [0,1] at random, and the probability of picking the same real twice is always zero (the last statement is only self consistently true, it's true in a countable model).

Making this precise is the job of forcing, which, given the political situation in mathematics, phrases everything in terms of the picking process, although at the end it talks about "random reals". But a random real number is an obviously consistent idea, even though it is incompatible with choice.

So when I look at post-forcing axioms like determinacy, I am always looking for the probability interpretation, and in this case I couldn't see it. The determinacy axiom, as I heard it, was exciting because it had a completely different character--- it asserted something about infinite sets that somebody thought was intuitively consistent (I don't know why, but they turned out to be right, so they must have had a good idea). But the justification for this for those with a formal view of higher infinity is that a large cardinal axiom proved that it is equiconsistent with set theory. So determinacy, as far as theorems about integers are concerned, is a moderate large cardinal axiom, and is not as interesting as a completely new axiom (although Cohen's "article of faith" says there are not going to be any consistent new axioms that are not equiconsistent by virtue of a large enough cardinal). So a superficial glance at Woodin's article was disappointing for me, because it didn't give a perspective which was probabalistic. But that's a very self-centered point of view, so I'll give it another shot.Likebox (talk) 17:51, 26 August 2008 (UTC)

## Need references

Edited out a minor error: ZFC stands for "Zermelo-Frankel with the axiom of choice", whereas ZF is simply "Zermelo-frankel". Would be nice to see some superscript references in this article, I'm afraid I don't know how to do that yet. 83.71.3.220 (talk) 00:41, 1 October 2008 (UTC)

See Wikipedia:Citing sources#Footnote summary for instructions on how to put in references. JRSpriggs (talk) 01:39, 1 October 2008 (UTC)

## Natural numbers?

Cantor introduced the concept of cardinality to compare the sizes of infinite sets, and he gave two proofs that the cardinality of the set of integers is strictly smaller than that of the set of real numbers. His proofs, however, give no indication of the extent to which the cardinality of the natural numbers is less than that of the real numbers.

Shouldn't natural numbers be rational numbers? Since the natural numbers is a subset of the integers, it's quite obvious that the cardinality of the natural numbers isn't greater than the integers. —Preceding unsigned comment added by 131.215.42.208 (talk) 23:21, 20 January 2009 (UTC)

The passage seems to be written for readers who know without having to think about it that the cardinality of the naturals is the same as that of the integers (or, possibly, who use the word integer to mean natural number, a convention used informally by some set theorists). Since not all readers will fall into this category, we should probably change both occurrences to be the same. The question is, should we say integers, or naturals? Integers sounds better (the repetition of natural numbers could be a bit stilted) but natural numbers is more likely to be strictly historically accurate when talking about what Cantor strictly speaking proved. --Trovatore (talk) 23:48, 20 January 2009 (UTC)

## What could be 2^aleph_0 without the CH?

I think the article should include a section saying which alephs could be ${\displaystyle 2^{\aleph _{0}}\,}$ (and ${\displaystyle 2^{\aleph _{\alpha }}\,}$) in the case CH is undecided. The article about the cardinality of the continuum has some indications about this (for example, it could be that ${\displaystyle 2^{\aleph _{0}}=\aleph _{\omega +42}\,}$, but not ${\displaystyle 2^{\aleph _{0}}=\aleph _{\omega }\,}$), but I think this article here is the proper place to a full analysis. And I have never seen an upper limit to ${\displaystyle 2^{\aleph _{0}}\,}$; does it make sense if ${\displaystyle 2^{\aleph _{0}}\,}$ could be some monstrosity like ${\displaystyle \aleph _{\omega _{\omega _{\omega }}+42}\,}$? Albmont (talk) 19:11, 3 April 2009 (UTC)

Well, for a realist there is only one thing the continuum could be, which is precisely whatever it is.
But if you're asking what values are consistent with ZFC, that's precisely known: It can be any regular cardinal, or any singular cardinal with uncountable cofinality. So yes, the value ${\displaystyle \aleph _{\omega _{\omega _{\omega }}+42}\,}$ is consistent with ZFC.
Of course one has to be a little careful how one states this. For example ${\displaystyle \left(2^{\aleph _{0}}\right)^{+}}$ is a regular cardinal, and yet it is obviously not consistent with ZFC that ${\displaystyle 2^{\aleph _{0}}=\left(2^{\aleph _{0}}\right)^{+}}$. I won't try to resolve this for you in this post; I'll leave you to puzzle over it a little while. --Trovatore (talk) 19:22, 3 April 2009 (UTC)
Since the article also talks about the denial of CH, I think it's proper to include in it the things that ${\displaystyle 2^{\aleph _{0}}\,}$ could be :-) Or, if we want to be precise, the article should list, within ZFC, which are the alphas for which we can prove that ${\displaystyle \aleph _{\alpha }\neq 2^{\aleph _{0}}\,}$
For example, I think that it's impossible to prove that ${\displaystyle 2^{\aleph _{0}}\neq \aleph _{\alpha +n}\,}$ for any α < ω1 and n > 0. But what about ${\displaystyle 2^{\aleph _{0}}\neq \omega 2\,}$? ${\displaystyle 2^{\aleph _{0}}\neq \omega \times \alpha \,}$? (BTW: maybe the article about cofinality should include a relation of the cf class-function to the +, . and ^ class-operations in the ordinal class).
I take it that ${\displaystyle 2^{\aleph _{0}}\,}$ can't (obviously) be its own successor, but can we prove that ${\displaystyle 2^{\aleph _{0}}\neq \aleph _{2^{\aleph _{0}}}\,}$ or even ${\displaystyle 2^{\aleph _{0}}\neq \aleph _{2^{2^{\aleph _{0}}}}\,}$? Albmont (talk) 20:21, 3 April 2009 (UTC)
I don't know that you've quite come to terms with the issue I was alluding to here. The problem with questions like "for what ${\displaystyle \alpha \,\!}$ can we prove that the continuum is not ${\displaystyle \aleph _{\alpha }}$?" is that you're mixing object language and metalanguage. Questions about provability are in the metalanguage, whereas "for what ${\displaystyle \alpha \,\!}$?" is in object language.
An example problem (not by any means the only problem): Some ordinals ${\displaystyle \alpha \,\!}$ (even some ordinals less than ω1 are not definable in the language of set theory. So in that case what does it mean to prove that the continuum is not ${\displaystyle \aleph _{\alpha }}$? By the most direct reading, you can't even get started, because you can't even state the claim that you're trying to prove. (And nevertheless, some of these ordinals are excluded, because they have cofinality ω, so we can't dismiss the issue either — it seems to mean something, but it's not obvious just what.)
There are reasonable formulations that take care of the problem, but as I say, you have to be careful. I think you need to think about it some more before anything I say about it will make much sense. --Trovatore (talk) 23:49, 3 April 2009 (UTC)
I think proper formalism is that, given a model of ZFC, and an ordinal ${\displaystyle \alpha }$ which is either a successor ordinal or has cofinality greater than ${\displaystyle \omega }$, then there is a model with the same ordinals and same cofinalities such that ${\displaystyle 2^{\aleph _{0}}=\aleph _{\alpha }}$. (I'm not sure that's accurate, but it's close.) However, it is not consistent that ${\displaystyle 2^{\aleph _{0}}=\aleph _{\alpha }}$ and ${\displaystyle \alpha }$ has countable cofinality. — Arthur Rubin (talk) 00:47, 4 April 2009 (UTC)
As for ${\displaystyle 2^{\aleph _{0}}=\aleph _{2^{\aleph _{0}}}\,}$ or even ${\displaystyle 2^{\aleph _{0}}=\aleph _{2^{2^{\aleph _{0}}}}\,}$, it doesn't make sense, as subscripts of ℵ are ordinals, and ${\displaystyle 2^{\aleph _{0}}}$ is a cardinal. If you use it to mean the initial ordinal corresponding to that cardinal, the first is consistent, but the second is not, by Cantor's theorem. The proof of the first is left as an exercise for the reader. — Arthur Rubin (talk) 00:52, 4 April 2009 (UTC)
Well, I know the answer, Arthur. I was trying to get Albamont to think through the definition-versus-denotation issue here. Until you do that, no answer to this question really makes sense.
Your formulation is indeed "close", but it has some problems. For example there isn't any (real) model, having the real ordinals, in which the continuum is larger than it is in V, because there aren't any more real numbers to add. There is a model in the sense of a forcing extension, but that's a different thing. --Trovatore (talk) 03:38, 4 April 2009 (UTC)
The reals that were added are pretty clearly new reals that just happen to not be in the smaller model... The difficulty is in the existence of generic filters over V. If you want, you can insist there are none. However, I've been told that the use of countable transitive models is out of vogue and forcing over V is "the way things are done" these days. — Carl (CBM · talk) 03:17, 6 April 2009 (UTC)
Forcing over V is indeed "the way things are done these days" — but that doesn't mean that any generic filters over V (for a nontrivial forcing notion) actually exist. You don't get any coherent ontology from the idea that they exist. Or at least I haven't been able to work out any way that you can, and I have tried. It seems that that path forces you to change your story — you have to say, look, first I took all the reals that there are (which you have to do, if you take the von Neumann hierarchy seriously), but now, lookee here, I found some more.
The right way to resolve this is, do whatever you have to do to find the right theorems (just as the Newton—Leibniz approach is still the best way of thinking of calculus), but realize that what you're actually talking about, when you force over V, is names for objects that don't really exist. When you ask, for example, whether one of these "new reals" is between 0.3 and 0.4, sometimes you're going to get the answer "maybe". No genuine real has that property. --Trovatore (talk) 22:41, 6 April 2009 (UTC)
Will respond on your talk page, since this is getting off topic. — Carl (CBM · talk) 00:56, 7 April 2009 (UTC)

You might want to look at Easton's theorem which deals with a more general question. JRSpriggs (talk) 06:37, 4 April 2009 (UTC)

That page seems to have the same problem I was talking to Arthur about — doesn't distinguish between honest-to-God models that live in V, and models for which you have to go to forcing extensions or Boolean valued models. It's true that it would be awkward to insert correct language for the point. --Trovatore (talk) 09:35, 4 April 2009 (UTC)

My question is quite simple: given ZFC, can we name a few values of α such that ${\displaystyle 2^{\aleph _{0}}\neq \aleph _{\alpha }\,}$? There's no need to expand the metalanguage to discuss models, etc - let's get to the basic, simple theory and see what we can get. For example, the first value that we can show is ${\displaystyle \alpha =\omega \,}$, the second is ${\displaystyle \alpha =\omega +\omega =\omega .2\,}$, etc. Of course, given that it is impossible to prove within ZFC that ${\displaystyle 2^{\aleph _{0}}\,}$ is ${\displaystyle \aleph _{\omega +42}\,}$ and others, the next question would be quite obvious, but let's not do that leap. Not yet. Albmont (talk) 03:13, 5 April 2009 (UTC)

I'm afraid your "simple question" simply doesn't make sense without some stipulations regarding the object language/metalanguage question. Now, it's not that a simple answer can't be given in some sense. The simple answer, below ω1 which I think you mentioned at some point, is that only the successor ordinals can be the α you're looking for.
But the problem is that that answer, though correct in some sense, is flat wrong unless you explain what you mean by it. Here's an example of why. Define α as follows:
α = 2 if CH holds
α = 1 otherwise
Now α is a successor ordinal below ℵ1, and yet it is inconsistent with ZFC that ${\displaystyle 2^{\aleph _{0}}=\aleph _{\alpha }}$.
So it's just incorrect to say that you can answer your question correctly without addressing this issue. --Trovatore (talk) 05:50, 5 April 2009 (UTC)
To Trovatore: Perhaps I am having trouble understanding your point, but the definition you gave for α is not a proper definition in my way of thinking. It is reminiscent of Grue and bleen. JRSpriggs (talk) 21:10, 5 April 2009 (UTC)
Why is it not a proper definition? It can be translated into a formula in the language of set theory, with one free variable, such that provably in ZFC, there is a unique object satisfying that formula, and that object has the properties I have ascribed to it. If that's not a proper definition, then you owe me some sort of demarcation between definitions that are proper and those that aren't. --Trovatore (talk) 02:00, 6 April 2009 (UTC)
Because of the failure of absoluteness here, one has to distinguish between the ordinal named by φ in the base model and the ordinal named by φ in the extension. For example, let φ be the formula above; if α is named by φ in the base model, there really will be another model in which ℵ(α) really is the cardinality of the continuum; however this will not be the ordinal defined by φ in the new model. Rather, one must look at the image of α under the embedding that is constructed when the new model is constructed. I don't know if this is worth pointing out somewhere; I believe it is usually taken for granted in the literature. — Carl (CBM · talk) 03:17, 6 April 2009 (UTC)

Albmont: the informal "short answer" is that any uncountable κ with uncountable cofinality is a possible candidate for the cardinality of the continuum. But this statement is not entirely precise because "possible" here means "can be achieved by a forcing extension" (because the informal statement is really just a much easier special case of Easton's theorem), and there are some technical issues related to forcing that have to be mastered in order to fully understand what's going on. These issues are what Trovatore is alluding to. — Carl (CBM · talk) 04:03, 6 April 2009 (UTC)

Thanks, Carl. Yes, absoluteness is the key. Since we are trying to talk about ordinals and cardinals in multiple models of set theory, we must use definitions which are independent of the choice of the model, i.e. absolute. So φ must be chosen to be Δ1 which I believe that Trovatore's definition of α is not. JRSpriggs (talk) 06:18, 6 April 2009 (UTC)
Certainly the α I defined is not absolute, and yes, that's one key. But the point is more general than that. Carl's "short answer", by itself, is simply underspecified.
People have to come to terms with the difference between an object and a definition of an object, or they will be led astray by this sort of discussion. It's something that most people never really come across — every natural number, for example, has kind of a canonical definition, and the distinction between object language and metalanguage doesn't really come up until you start talking about the Goedel theorems or something. But for ordinals there's just no way around it; you have to come to terms with the problem.
This is the same problem with discussions such as whether anyone can "find" a counterexample to CH, in the sense of a set of reals of intermediate cardinality. --Trovatore (talk) 09:23, 6 April 2009 (UTC)

I don't believe the "object language"/"metalanguage" distinction is really the problem here. In these sorts of situations we work model theoretically: start with a model W and work with a language that has a constant for every element of W. So every ordinal in W is definable in this language. One thing that does matter is to distinguish between things that look like names but are actually meant to be formulas, and things that look like names that are actually meant to be names. This is typically achieved by writing dots over the things that are meant to be names; we ought to write ${\displaystyle 2^{\aleph _{0}}={\dot {\alpha }}}$.

So a slightly better way of stating the theorem at hand goes like this.

Let W be a model of set theory and assume that certain generic sets over W exist. Then for any cardinal α in W which (in W) is uncountable with uncountable cofinality, there is a forcing extension which preserves cardinalities in which ${\displaystyle 2^{\aleph _{0}}={\dot {\alpha }}}$.

I don't see that it matters whether any formula φ defining α is absolute or not, because we will not actually use φ to identify a cardinal in the forcing extension, we use the forcing name. The absoluteness issue is only relevant to seeing why Trovatore's example does not give a contradiction to Easton's theorem. Easton's theorem should have a dot above the G(α), that's all. But I can see why it is usually not included. — Carl (CBM · talk) 12:13, 6 April 2009 (UTC)

The reason it's necessary to make the object language — metalanguage distinction here is that the actual question was not model-theoretic, but rather proof-theoretic. And the proof-theoretic question looks like it ought to make sense, but it actually doesn't. I think we need to make clear, when giving the "in some sense correct" answer, that it is not actually an answer to the (fundamentally ill-formed) proof-theoretic question.
Of course it does answer certain instances of the proof-theoretic question, after reformulation to make them well-formed, but the person who wants to understand this really needs to go learn forcing. Once he or she learns forcing he can figure out these instances and reformulations for himself. --Trovatore (talk) 17:44, 8 April 2009 (UTC)
I think we read the initial question differently. — Carl (CBM · talk) 20:00, 8 April 2009 (UTC)
Does the original question make sense if it is reformulated this way: Are there any ordinals ${\displaystyle \alpha }$ for which it has been shown that ${\displaystyle 2^{\aleph _{0}}\ \neq \aleph _{\alpha }}$? If yes, could the article list at least one of them? Or is it just wrong to state ${\displaystyle 2^{\aleph _{0}}\neq \aleph _{\omega }}$ without supplyning a PHD theseis?YohanN7 (talk) 21:43, 10 April 2009 (UTC)
It makes less sense that way; you are running into the issue Trovatore was concerned about. I think that the answer you are looking for is the following fact. There is no model of ZFC in which the equation ${\displaystyle 2^{\aleph _{0}}=\aleph _{\omega }}$ holds. — Carl (CBM · talk) 21:47, 10 April 2009 (UTC)
Ok, I'll read this discussion more carefully and try to understand. B t w, I meant in my question within ZFC (and the negation CH of course).YohanN7 (talk) 21:57, 10 April 2009 (UTC)
To give a simplistic answer — ${\displaystyle 2^{\aleph _{0}}\,}$ must obey ${\displaystyle \omega <\operatorname {cf} (2^{\aleph _{0}})\,.}$ So it cannot be: finite, ${\displaystyle \aleph _{0}\,,}$ ${\displaystyle \aleph _{\omega }\,,}$ ${\displaystyle \aleph _{\omega \cdot 2}\,,}$ ${\displaystyle \aleph _{\omega \cdot 3}\,,}$ ... ${\displaystyle \aleph _{\omega ^{2}}\,,}$ ... etc. That is, most singular cardinals are not allowed, but ${\displaystyle \aleph _{\omega _{1}}\,}$ would be OK because its cofinality is ${\displaystyle \omega _{1}>\omega \,.}$
On the other hand most regular cardinals would be OK (could be ${\displaystyle 2^{\aleph _{0}}\,}$), specifically: ${\displaystyle \aleph _{1}\,,\aleph _{2}\,,\aleph _{3}\,,\aleph _{4}\,,\aleph _{5}\,,\ldots \,}$${\displaystyle \aleph _{\omega +1}\,,\aleph _{\omega +2}\,,\ldots \,.}$ OK? JRSpriggs (talk) 12:08, 11 April 2009 (UTC)
So, here came my next question (that apparently has a negative answer). It it possible to give an explicit ordinal α for which it can be shown that ${\displaystyle 2^{\aleph _{0}}<\aleph _{\alpha }\,}$? Albmont (talk) 12:00, 16 April 2009 (UTC)
BTW, if I got it right, PCF theory provides an upper bound for ${\displaystyle 2^{\aleph _{0}}\,}$. Albmont (talk) 12:15, 16 April 2009 (UTC)
Certainly it is possible to give an explict ordinal α and prove ${\displaystyle 2^{\aleph _{0}}<\aleph _{\alpha }}$. Just let ${\displaystyle \alpha =\left(2^{\aleph _{0}}\right)+1}$. I think you misread the bit about PCF theory, though. --Trovatore (talk) 17:40, 16 April 2009 (UTC)
This would depend, of course, on whether one accepts that ${\displaystyle \alpha =}$ actually refers to a specific ordinal. But I agree that Albmont's question here mixes metalanguage and object language, and that the only way to really understand what is going on is going to be to learn more about forcing in more detail. — Carl (CBM · talk) 23:16, 16 April 2009 (UTC)
Independently of whether you think that it represents a specific ordinal (as opposed to what? a nonspecific ordinal?), it is undeniably the case that ZFC proves the formula ${\displaystyle \alpha =\left(2^{\aleph _{0}}\right)+1\implies 2^{\aleph _{0}}<\aleph _{\alpha }}$. The problem, besides the object–meta thing, or possibly as a rephrasing of it, is that the word "explicit" is doing unexamined work here. --Trovatore (talk) 23:28, 16 April 2009 (UTC)
Yes, of course ZFC proves that formula. The problem with mixing "explicit" and "proves" is that the former is best handled in a language with a constant for every ordinal, while the latter is usually handled in the language without those constants. — Carl (CBM · talk) 00:06, 17 April 2009 (UTC)
Albmont, on the PCF thing: My guess is that you looked at the result that said that ${\displaystyle 2^{\aleph _{\omega }}<\aleph _{\aleph _{4}}}$, granted that ${\displaystyle \aleph _{\omega }}$ is a strong limit cardinal, and reasoned that since ${\displaystyle 2^{\aleph _{0}}\leq 2^{\aleph _{\omega }}}$, it follows that ${\displaystyle 2^{\aleph _{0}}<\aleph _{\aleph _{4}}}$. That would be fine, except that it ignores the "granted that..." part.
Once you know (or assume) that ${\displaystyle \aleph _{\omega }}$ is a strong limit, you actually don't need PCF theory to prove that ${\displaystyle 2^{\aleph _{0}}<\aleph _{\aleph _{4}}}$. If it's not obvious, think about it for a minute. When you see it you'll go ohh. --Trovatore (talk) 09:17, 17 April 2009 (UTC)
Yes, I saw that inequality, but I also saw the "granted that..." part - which hindered me to write ${\displaystyle 2^{\aleph _{0}}<\aleph _{\omega _{4}}\,}$ (a note on style: is it ok to write ${\displaystyle \aleph _{x}\,}$, where x is a cardinal?). Does the PCF thing give any upper limit without any extra axioms beyond ZFC? Unfortunately, I think it will take more than a minute to think about it :-) Albmont (talk) 17:14, 17 April 2009 (UTC)
pcf theory will give you no bounds on the continuum. pcf theory is about powers of singular cardinals. It is certainly possible that the continuum is ${\displaystyle \aleph _{\omega _{5}}}$. ${\displaystyle 2^{\aleph _{0}}}$ can be anything it ought to be. (Robert Solovay, 1965) --Aleph4 (talk) 18:23, 17 April 2009 (UTC)
To Albmont: I am not familiar with PCF theory, but the article on it says "It gives strong upper bounds on the cardinalities of power sets of singular cardinals.". I would just point out that ${\displaystyle \aleph _{0}}$ is regular, not singular. Thus it appears that the theory does not apply to it. Anyway, Easton's theorem implies that there is no upper bound on ${\displaystyle 2^{\aleph _{0}}\,.}$ JRSpriggs (talk) 20:01, 17 April 2009 (UTC)
Well, ${\displaystyle 2^{\aleph _{0}}}$ is necessarily the smallest cardinality of a powerset of an infinite set, so any bound on the powerset of a singular cardinal must ipso facto bound the continuum.
Spoiling my own riddle above, PCF theory does indeed imply that, if ${\displaystyle \aleph _{\omega }}$ is a strong limit, then the continuum is less than ${\displaystyle \aleph _{\aleph _{4}}}$. But this is trivial — by the definition of "strong limit", since ${\displaystyle \aleph _{0}<\aleph _{\omega }}$, it's immediate, under the given assumption, that ${\displaystyle 2^{\aleph _{0}}<\aleph _{\omega }}$. --Trovatore (talk) 07:44, 18 April 2009 (UTC)

## CH in a non-AC context

Yuck. This is the kind of issue I'd ordinarily prefer to just ignore. But the issue has been raised at axiom of determinacy and seems likely to recur.

The question is, which definition of CH should we give as primary, the one I call weak CH, which says every uncountable set of reals is equinumerous with the reals, or strong CH, which says the reals are equinumerous with the countable ordinals (that is, ${\displaystyle 2^{\aleph _{0}}=\aleph _{1}}$)?

It's true that most sources seem to state it in a weak form (I'm counting there is no cardinality strictly between ${\displaystyle \aleph _{0}}$ and ${\displaystyle 2^{\aleph _{0}}}$ as an instance of the weak form; it's equivalent modulo the violent pathology of an infinite Dedekind-finite set of reals). Jech is a notable exception; he explicitly equates the continuum hypothesis with ${\displaystyle 2^{\aleph _{0}}=\aleph _{1}}$.

However, in the default set-theoretic context, the axiom of choice holds, so there is no need to make the distinction. Therefore I think it is unjustified to assume that this choice of statement reflects a preference as to the essential meaning of CH in a non-AC context. I would actually argue that strong CH is more natural to think of as capturing that meaning from the standpoint of contemporary set theory; I've given a couple of reasons at talk:axiom of determinacy.

The problem at the AD article is that an editor wanted to add the claim that AD implies CH, which I think is severely contrary to the usage of the terminology in the set-theoretic community. I've worked with set theorists who study AD, and I never recall any of them putting it that way. The problem for this article is that it puts the weak CH statement first, and then asserts that, given AC, this is equivalent to strong CH; this I think is misleading. Better would be to state them both, note that they're equivalent, but then note that in the absence of AC, they're not. --Trovatore (talk) 19:59, 24 April 2009 (UTC)

1. "weak CH" is sometimes used for ${\displaystyle 2^{\aleph _{0}}<2^{\aleph _{1}}}$. But let's keep your terminology for the moment.
2. I think that in the non-AC context, "weak CH" is more interesting, because strong CH rarely (in the sociological sense) holds.
3. The "most sources" that you mention include Cantor and Hilbert (in his first problem).
4. There is a theorem of Sierpinski that GCH implies CH. This theorem is about weak GCH; the implication from strong GCH is less interesting, since it factors through the property "every power set of an ordinal can be well-ordered", which is not about cardinal arithmetic at all.
Therefore I would prefer weak CH as the "primary" definition of CH. (Despite Jech. We also don't follow Kunen in phrasing AC as "every set can be well-ordered".) As far as the AD article is concerned, I think the current version (with the inline explanation of precisely which form of CH it implies) is quite clear.
--Aleph4 (talk) 16:12, 8 May 2009 (UTC)
Although 4 is an interesting point, I still don't agree:

## Kurt Godel's Proof

The section "As the first Hilbert problem" and "Impossibility of proof and disproof (in ZFC)" don't agree on Kurt Godel's work using the AC:

Later work by Kurt Gödel in 1939 showed that the continuum hypothesis could not be disproved based on the current axioms of set theory (ZF).

vs

Kurt Gödel showed in 1940 that the continuum hypothesis (CH for short) cannot be disproved from the standard Zermelo-Fraenkel set theory (ZF), even if the axiom of choice is adopted (ZFC).

The former only mentions that Godel did his work in ZF, but the latter section says that his original work was under the ZFC. Given the results, I think it was under ZFC. I think the former section needs to be made more explicit. --B-Con (talk) 08:11, 15 May 2009 (UTC)

Gödel showed that ~CH cannot be proved in ZF and also cannot be proved in ZFC. — Carl (CBM · talk) 11:37, 15 May 2009 (UTC)
To Carl: I think you mean "disproved" rather than "proved". Do not confuse him with Paul Cohen (mathematician).
To B-Con: Kurt Gödel showed that within a model of ZF, one can define the constructible universe (L) a proper class which not only satisfies ZF but also GCH and thus both CH and the axiom of choice. Thus, if CH could be disproved from ZF or from ZFC, then one would have reached a contradiction in this submodel and thus a contradiction in ZF itself. JRSpriggs (talk) 10:10, 16 May 2009 (UTC)
Thanks, I meant the negation of CH of course, and so I added a tilde above. Re B-Con: it's better to look things up, or ask, rather than making guesses. — Carl (CBM · talk) 15:14, 16 May 2009 (UTC)

## Cantor's fanatism

I was just reading a book about Cantor. It mentioned that one of the main reasons Cantor believed in the continuum hypothesis was because he managed to show every nonempty perfect set is the same size as the reals, and then showed a closed set C is either countable or has cardinality of the reals by partitioning C into a perfect set and a countable set. And then he said "In future paragraphs it will be proven that this remarkable theorem has a further validity even for linear point sets which are not closed,..." Should this be in the article? Also I personally believe his reasoning is wrong: since the reals have a countable base, the number of open sets is the same as the reals, and thus so is the number of closed sets (because they are complements of open sets). But there are "so many more" subsets, because 2^(aleph0) is negligible compared to 2^(2^(aleph0)). Right???Standard Oil (talk) 14:28, 14 August 2009 (UTC)

Cantor never thought he had proved CH, iirc. Yes, a quote from him (and a little more info on the origins of the problem--Cantor did a lot of work on it) would be nice. 66.127.52.47 (talk) 11:32, 24 March 2010 (UTC)

## Goedel

Could someone clueful please check whether this edit (by me) is any good, and revert it if necessary. I started having doubts after making it. Thanks. 66.127.55.192 (talk) 17:57, 17 February 2010 (UTC)

Sounds better to me. JRSpriggs (talk) 04:19, 18 February 2010 (UTC)

## Dehornoy article

an article explaining(?) Woodin's recent work. I don't understand it at all. Maybe someone here can make sense of it and possibly cite it or use something from it. 66.127.55.192 (talk) 06:24, 18 February 2010 (UTC)

## Why the name?

Why is this hypothesis called the "continuum" hypothesis? The proposition "There is no set whose cardinality is strictly between that of the integers and that of the real numbers." is asserting the lack of any set of size (or cardinality) between integers and real numbers, which sounds far more like the a "non-continuum hypothesis." or since it is saying that there is not a single set that stands in that range (far less a continuum of size/cardinalities of sets) the theory might even be called the "not-even-close-to-continuum hypothesis" or "discontinuum hypothesis" or "discrete hypothesis." Since continuum is a word in normal English, I guess it would help to know why this theory has the name it has. I can't understand the name, let alone the math :-) --Timtak (talk) 10:09, 14 October 2010 (UTC)

Answering my own question, looking at the article on Continuum_(set_theory), the latter seems to be saying that the fact that there are no sets between the size (cardinality) of integers and real numbers, proves that the latter forms a continuum? I.e. the continuum hypothesis, is called such, not because it says something about a continuum of sets, but because it posits a continuum of real numbers. So perhaps by way of explanation, one might write, "'There is no set whose cardinality is strictly between that of the integers and that of the real numbers.' implies that real numbers form a continuum, and hence the name of this theory." Is that it? Or indeed, since from the the Continuum_(set_theory) article it says that the cardinality of real numbers is the cardinality of the real line (is that the same as "a or any real line"?) then the continuum hypothesis is called such because it implies reality is continuous?!--Timtak (talk) 10:18, 14 October 2010 (UTC)

It's classical terminology. The real numbers were traditionally called "the continuum" and the continuum hypothesis is a hypothesis about the cardinality of the real numbers. — Carl (CBM · talk) 10:51, 14 October 2010 (UTC)

Perhaps because it could be worded "Any uncountable subset of the continuum is equinumerous with the continuum.". JRSpriggs (talk) 11:55, 14 October 2010 (UTC)

From my classes we were taught that [0,2\pi] was called "the continuum", (which of course has the same cardinality as the Reals), but nevertheless Cantor was interested in Fourier series convergence, and all those fellas ever talk about is [0,2\pi] :P . Handsofftibet (talk) 13:15, 13 January 2011 (UTC)

Where was your teacher trained? Tkuvho (talk) 13:32, 13 January 2011 (UTC)

Australia and University of Newcastle Upon Tyne - EnglandHandsofftibet (talk) 11:35, 14 February 2011 (UTC)

Truncating the continuum at 0 and 2\pi seems a bit unusual. I have certainly not seen any trace of such a convention at any of the wiki pages on varieties of continua. Also, historically mathematicians have studied various notions of the continuum but it always tends to be a linear one, see continuum (theory). Tkuvho (talk) 17:31, 14 February 2011 (UTC)

## No inverse powerset

Template:User-multi and Template:User-multi added a reference to "Deciding the Continuum Hypothesis with the Inverse Power Set" by Patrick St-Amant. This paper begins by adding a new axiom to ZFC, namely

${\displaystyle \forall X\exists Y(P(Y)=X)\,.}$

This axiom is inconsistent with the existence of the empty set. The existence of the empty set is an unavoidable consequence of the existence of any set and the axiom of separation. The empty set is defined by

${\displaystyle \forall A(\lnot \,A\in \varnothing )\,.}$

The powerset of a set B is defined by

${\displaystyle \forall C(\forall D(D\in C\rightarrow D\in B)\leftrightarrow C\in P(B))\,.}$

If the new axiom were true, then we could instantiate it for the empty set to give

${\displaystyle P(Y)=\varnothing \,}$

for some set Y. Then putting Y in for B in the definition of powerset gives

${\displaystyle \forall C(\forall D(D\in C\rightarrow D\in Y)\leftrightarrow C\in \varnothing )\,.}$

In particular, if C were the emptyset, we would get

${\displaystyle \forall D(D\in \varnothing \rightarrow D\in Y)\leftrightarrow \varnothing \in \varnothing \,.}$

By the definition of the emptyset

${\displaystyle \lnot D\in \varnothing \,}$

for any D, and thus

${\displaystyle \forall D(D\in \varnothing \rightarrow D\in Y)\,}$

holds vacously. Consequently, we would arrive at

${\displaystyle \varnothing \in \varnothing \,}$

which contradicts the definition of the empty set. Since the new "axiom" led to a contradiction (using only definitions and the existence of the empty set), it must be false. Thus Patrick St-Amant's whole project is an exercise in futility, in fact, an obvious hoax. JRSpriggs (talk) 12:29, 6 January 2011 (UTC)

It might not be, actually. St-Amant seems to be proposing a kind of a Grothendieck construction to "extend" the universe of sets to a kind of an object of negative rank in the set-theoretic universe. Note that he seems to have a paper published in a refereed journal. The text does not seem funny enough to be merely a hoax. But at any rate it is far from clear that it meets notability guidelines. Tkuvho (talk) 13:48, 6 January 2011 (UTC)
Hoaxes need not be "funny", i.e. comical. They may be serious attempts to deceive people, perhaps with the intention of putting the victims down in order to make the perpetrator seem superior by comparison.
If the universe of sets is to be extended, then he needs to make it clear at the outset what rules of ZFC are being relaxed (negated or altered), instead of pretending to preserve them all. In this case, how has he changed the definitions of the empty set, of powerset, or the axiom of separation? Otherwise the argument above will still apply and render his work meaningless. JRSpriggs (talk) 21:05, 6 January 2011 (UTC)
By the way, I should note that: when a set X is the powerset of a set Y, Y is the union of X. So a one-sided inverse already exists. What we are arguing about here is whether there could be inverse on the other side of the powerset operation. Ironically, St-Amant purports to define his inverse operation in his second axiom, but he uses the wrong side relative to what he said in his first axiom. Thus the second axiom could be construed as referring to the union operation. JRSpriggs (talk) 06:00, 7 January 2011 (UTC)
I imagine he means the usual axioms to apply to "sets" whose rank is a natural number, or something like that. I you sure he did not say something to that effect somewhere in the 30 page paper? I don't really have the set-theoretic wherewithal to understand the paper, but why are we trying so hard to refute it? An alternative approach would be to encourage this new contributor to add this material once it is published and its notability is established. Tkuvho (talk) 06:03, 7 January 2011 (UTC)
I am not going to waste my time trying to read and understand a long paper when I find a fundamental error on the second page. The reason I wrote this section of talk to refute it is that at Wikipedia talk:WikiProject Mathematics#Hoax warning some people expressed doubts about my claim that this is a hoax. JRSpriggs (talk) 06:10, 7 January 2011 (UTC)
I would assume that before one claims something is a hoax, one would necessarily have to waste his time at least perusing the paper in question. On the second page you mentioned, the author clearly states that the power axiom is supposed to apply only to a certain class of objects in his theory, whereas the other axioms of ZF are assumed to apply only to "sets which arise from ZF". Tkuvho (talk) 11:45, 7 January 2011 (UTC)
We're applying the wrong standard for removal. It may not be a hoax, but it's not well-formed (it doesn't describe the complete axiom systems, so that proofs could be verified, nor is there evidence the resulting system would be consistent) and there is no sign of reliable sources referring to it. It should not appear in a Wikipedia article, even as an external link, until someone reliable talks about it. — Arthur Rubin (talk) 15:36, 7 January 2011 (UTC)
That the author is published doesn't make his unpublished papers reliable. He would have to be an expert in the field; a minimum qualification would be that he has published papers in the field. The range of fields of his unpublished papers argued about in Wikipedia articles suggests he is not an expert. — Arthur Rubin (talk) 15:42, 7 January 2011 (UTC)

Template:Unindent To Tkuvko: If you understand his theory, then tell me — Where am I wrong in my deduction of a contradiction from his theory? Which is the first step that fails and why? JRSpriggs (talk) 11:08, 8 January 2011 (UTC)

Since the empty set is not the powerset of anything, the "new axiom" fails on first use. -- cheers, Michael C. Price talk 14:28, 8 January 2011 (UTC)
Look I am not acquianted with St Amant and have no vested interest in defending his theory, though I am curious about it. Perhaps your argument fails in the second step. Why do you assume that his idealized power set operation is supposed to be surjective onto traditional sets? Tkuvho (talk) 16:48, 8 January 2011 (UTC)
To Tkuvho: If I understand you, you are saying that my inference from
${\displaystyle \forall X\exists Y(P(Y)=X)\,}$
to its instance
${\displaystyle \exists Y(P(Y)=\varnothing )\,}$
might be wrong. That is, St-Amant may not intend that all normal sets (which he calls Zermelo sets) be covered by his axiom 1. However, he does give several examples of normal sets to which he applies P−1, some of which do not contain the empty set as an element (which is all I really need). Can you quote any sentence in his article which gives a criterion for excluding some normal sets from the domain of P−1? JRSpriggs (talk) 23:09, 9 January 2011 (UTC)
Well I would have hoped one of the 70.etc IP's might have given us an explanation if treated in a more welcoming fashion. Tkuvho (talk) 01:13, 10 January 2011 (UTC)

## Large cardinal?

Is it consistent with ZFC to have ${\displaystyle 2^{\aleph _{0}}}$ = some large cardinal?YohanN7 (talk) 07:48, 9 May 2011 (UTC)

Well, not the "normal" ones, so to speak. But there are variants that still fit neatly into the consistency-strength hierarchy for which the answer could be yes. For example, it's consistent that the continuum is weakly inaccessible (equiconsistent with the existence of an inaccessible) or that it's real-valued measurable (equiconsistent with the existence of a measurable). --Trovatore (talk) 07:53, 9 May 2011 (UTC)
The two cases you mention are related in a way, right? I vaguely recall reading in "Geometric Measure Theory" by Herbert Federer (a horribly difficult book for me by the way, but good) that the existence of a nontrivially measurable infinite set requires the existence of an inaccesible cardinal. I need to refresh my brief knowledge on that. By the way, the articles here on inaccessible cardinals and related definitions are kind of messy. (I can be more precise regarding that last statement if needed, given some time.)YohanN7 (talk) 09:12, 9 May 2011 (UTC)
Edit: In the above I meant measurable set with all subsets measurable. YohanN7 (talk) 19:03, 9 May 2011 (UTC)
Right, I believe that requires a real-valued measurable, but I'd have to check myself on that. Real-valued measurables are certainly all weakly inaccessible, and quite a bit more than that (e.g. weakly Mahlo). I have heard of something called a "real-valued supercompact", so it doesn't necessarily stop at measurable. I wonder if there's some abstract way for taking a large-cardinal property, defined say in terms of extenders, and turning all the measures into real-valued measures, and thereby getting a "real-valued" version. --Trovatore (talk) 19:08, 9 May 2011 (UTC)
As far as the measures go, signed measures and complex-valued measures can be expressed in terms of real-valued measures (positive semidefinite ones) using the Hahn-Jordan decomposition. But you probably knew that. [I'm way out of my league here;)]YohanN7 (talk) 21:04, 9 May 2011 (UTC)
Ah, just figured out what you were talking about here. That's not really what I meant. The measures given by extenders, when defining large cardinals, are ordinarily two-valued measures — the only possible values are 0 and 1. Going to real-valued measures is a liberalization, not an extra requirement. --Trovatore (talk) 03:07, 10 May 2011 (UTC)
Hm. I should perhaps have read real-valued measurable before I opened my mouth;). YohanN7 (talk) 09:52, 10 May 2011 (UTC)
Indeed, we are talking about the same thing, at least we were at the outset. My reference [Federer] refers to non-measurable cardinals as Ulam Numbers. The only thing that appears to differ is the term of atomless that gives the transition from measurable to real-valued measurable (implicitely there in Federer I believe). Do you mean that you want to take an arbitrary measure and expand it into some sort of sum of real-valued measures? And, in addition, somehow express various large cardinal properties to see exacly how strong large cardinal axiom one needs to actually find a cardinal that at least might be real-valued measurable? In other words, you want to find exactly where in the supposedly linearly ordered consistency-strength hierarcy of large cardinal axioms each of the involved axioms belong? (I'm only guessing here so don't laugh too much!) YohanN7 (talk) 10:28, 10 May 2011 (UTC)
Easton's theorem allows one to make ${\displaystyle 2^{\aleph _{0}}}$ equal to be any uncountable regular cardinal since they all have cofinality greater than ${\displaystyle \omega \,.}$ However, when you do that the cardinal in question would cease to be a strong limit cardinal and thus would no longer satisfy the definitions of most types of large cardinals. JRSpriggs (talk) 10:03, 9 May 2011 (UTC)
Yes, that makes sense. But you must mean that the cardinal in question would cease to be a strongly inaccessible cardinal (if that is what we happened to start with)? Might it still be a weakly inaccessible cardinal? YohanN7 (talk) 22:28, 9 May 2011 (UTC)
If you do it in the most obvious way (just add κ Cohen reals) then I think the forcing preserves cardinalities and cofinalities, so κ should still be both a limit cardinal and a regular cardinal, thus weakly inaccessible. --Trovatore (talk) 01:06, 10 May 2011 (UTC)

## The Axiom of symmetry

One intuitive argument against CH is said to be Freiling's Axiom of Symmetry. I wonder where the intuition comes from. I agree that the Axiom of Symmetry (AX) sounds quite convincing on first reading, but that's a little thin for mathematical reasoning. I thought about a way to base the intuition on provable facts using smaller sets than the reals. What about this?

Modify AX the following way:

1.) We throw darts at a countable set instead of the unit interval (I). Say the natural numbers (N) for definiteness.

2.) The function f:I->{x:x is a countable subset of I} occuring in AX is exchanged for f:N->{x:x is a finite subset of N}

Leave everything else untouched in AX. Call this thing the Axiom of Countable Symmetry (ACX).

I think that ACX "sounds" just about as convincing as AX on "first reading". Lets call a counterexample of AX (or ACX) a Freiling function. ACX is false. Proof: f(n) = {m: m <= n} is a Freiling function.

I don't draw any conclusions from this. But provided I'm correct in the above [and I might very well not be, the night is not young;)], I can't see the validity of the claim that AX is an argument against CH. If anything, it might be an argument in favour of CH, but I wouldn't go as far as that. In the article Freiling's axiom of symmetry there are listed two objections against AX. My example above might just be an example of objection number two, but I can't see exactly what that number two says. At any rate one does not need CH to disprove ACX, and I can't see the need for AC either. For a countable set there exists a bijection between it and N. That bijection does provide a wellordering (with the same length as N). Or doesn't it? Correct me if I am wrong. YohanN7 (talk) 00:00, 10 May 2011 (UTC)

That's the point call Indduction i and strong induction I

Then what I have proved it that ZFCI<=>ZFCi<=>ZFCiCH<=>ZFCCH hence following it back and forth ZFCI<=>ZFCCH Hence If we assume ZFC, and not induction, we have not CH.

Hence ZFCi<=>ZFCI<=>ZFCRH which implies trivially that ZFCi<=>ZFCRH you can just never right down all the maps from R to N and N to R at the same time, but you can approximate them by 1-1 function f_N:I->R and the function f(x)= x and f(x)=-x:R->R and hence I->I invertibaly.

you should probably understand now, thanks for looking :) — Preceding unsigned comment added by WhatisFGH (talkcontribs) 03:46, 20 October 2011 (UTC)

## Feferman/EFI

I added a note[1] about an article by Solomon Feferman, though I think I may have messed up the explanation somewhat—I figured the main thing was to get the citation into the article. Any review/fixes would be appreciated. The EFI page (url in the Feferman citation) looks interesting in general and there's an overview by Koellner[2] of the current state of CH, that also seems worth summarizing in the article. I might try, but I'm not very knowledgeable, so it would be great if someone else did it. Feferman has some other slides about his "definiteness" theory[3] that are a little more detailed than his EFI slides, if that's of any interest. 64.160.39.72 (talk) 01:57, 5 February 2012 (UTC)

## Woodin

Hugh Woodin appears to have changed his mind regarding the truth value of CH. This should be reflected in the article. (See e.g. the Hugh Woodin wiki page.) YohanN7 (talk) 22:35, 9 April 2012 (UTC)

Can you link a recent paper by him to that effect? JRSpriggs (talk) 11:16, 10 April 2012 (UTC)
No, I have nothing better than hearsay, meaning what you find on the net when searching "Ultimate L continuum hypothesis" in Google. I'll look a little more carefully...
Here is a nontechnical article. Better than nothing, but not by him, and thus not good enough for a reference. http://www.newscientist.com/article/mg21128231.400-ultimate-logic-to-infinity-and-beyond.html?full=true
YohanN7 (talk) 15:33, 10 April 2012 (UTC)

Never mind the article. It will not be changed before a publication comes anyway. The interesting thing for me is if he has changed his mind. Does anyone of you gurus have an idea? Ultimate L seems to be a model of ZFC + a bunch of large cardinal axioms (including Woodin cardinals). There are lecture notes on it from a set theory workshop in 2010 available on the net. YohanN7 (talk) 20:20, 10 April 2012 (UTC)

By secondhand conversation, I do believe he has changed his mind at least to some extent. His previous argument apparently rested on an argument that he thought had that would show there was a limit to the large cardinals you can have in HOD. When he really tried to nail this argument down, it apparently fell apart, and so he had to rethink. The details of what he thinks now, I couldn't tell you. --Trovatore (talk) 21:13, 10 April 2012 (UTC)
Tnx for reply Trovatore. The reason that I'm interested is that CH is such a great mathematical problem. It's deep deep deep, but still a layman like me can understand the statement of it. YohanN7 (talk) 11:29, 13 April 2012 (UTC)

## Questions

Does anyone know, if;

${\displaystyle 2^{\aleph _{\alpha }}=\aleph _{\alpha +1}.}$

What this evaluates to?

${\displaystyle 2^{{{2}^{2}}^{.}..}.}$

Edit: It won't let me format that how i wanted to - basically I mean what is 2 to the power of 2 to the power of 2 to the power of 2 to the power of 2 and so on. — Preceding unsigned comment added by 109.149.174.130 (talkcontribs) 16:24, 23 October 2012‎

Hi 109. This is an article talk page, which is intended for discussions aiming at improving the article. General questions like this can be asked at Wikipedia:Reference desk/Mathematics. --Trovatore (talk) 19:18, 23 October 2012 (UTC)

My apologies, cheers. — Preceding unsigned comment added by 86.151.16.138 (talk) 20:37, 28 October 2012 (UTC)

## Proofs of certain consequences of GCH

Since a recent edit has implicitly raised the question of how the formulas in Continuum hypothesis#Implications of GCH for cardinal exponentiation are justified. I will provide proofs here.

First, take notice of the following facts:

Now, suppose that α ≤ β+1, then

${\displaystyle \aleph _{\beta +1}=2^{\aleph _{\beta }}\leq \aleph _{\alpha }^{\aleph _{\beta }}\leq \aleph _{\beta +1}^{\aleph _{\beta }}=(2^{\aleph _{\beta }})^{\aleph _{\beta }}=2^{(\aleph _{\beta }\cdot \aleph _{\beta })}=2^{\aleph _{\beta }}=\aleph _{\beta +1}\,}$

which means

${\displaystyle \aleph _{\alpha }^{\aleph _{\beta }}=\aleph _{\beta +1}\,.}$

On the other hand, suppose that β+1 < α, then

${\displaystyle \aleph _{\alpha }=\aleph _{\alpha }^{1}\leq \aleph _{\alpha }^{\aleph _{\beta }}\leq (2^{\aleph _{\alpha }})^{\aleph _{\beta }}=2^{(\aleph _{\alpha }\cdot \aleph _{\beta })}=2^{\aleph _{\alpha }}=\aleph _{\alpha +1}\,}$

which means

${\displaystyle \aleph _{\alpha }\leq \aleph _{\alpha }^{\aleph _{\beta }}\leq \aleph _{\alpha +1}\,.}$

If we further suppose that ${\displaystyle \aleph _{\beta }<\operatorname {cf} (\aleph _{\alpha })}$ where cf is the cofinality operation, then any function from ${\displaystyle \aleph _{\beta }\,}$ to ${\displaystyle \aleph _{\alpha }\,}$ must be bounded above by some ${\displaystyle \gamma <\aleph _{\alpha }\,.}$ And γ has a cardinality ${\displaystyle \vert \gamma \vert =\aleph _{\delta }\,}$ where δ < α. The cardinality of the set of functions so bounded by γ is

${\displaystyle \aleph _{\delta }^{\aleph _{\beta }}\leq 2^{(\aleph _{\delta }\cdot \aleph _{\beta })}=\aleph _{\max(\delta ,\beta )+1}\leq \aleph _{\alpha }\,.}$

Adding these together for the ${\displaystyle \aleph _{\alpha }\,}$ possible values of γ gives

${\displaystyle \aleph _{\alpha }^{\aleph _{\beta }}\leq \aleph _{\alpha }\cdot \aleph _{\alpha }=\aleph _{\alpha }\,}$

which means

${\displaystyle \aleph _{\alpha }^{\aleph _{\beta }}=\aleph _{\alpha }\,.}$

On the other hand, if we further suppose ${\displaystyle \operatorname {cf} (\aleph _{\alpha })\leq \aleph _{\beta }\,,}$ then by one of the corollaries of König's theorem we have

${\displaystyle \aleph _{\alpha }<\aleph _{\alpha }^{\operatorname {cf} (\aleph _{\alpha })}\,}$

and thus

${\displaystyle \aleph _{\alpha +1}\leq \aleph _{\alpha }^{\operatorname {cf} (\aleph _{\alpha })}\leq \aleph _{\alpha }^{\aleph _{\beta }}\,}$

which means

${\displaystyle \aleph _{\alpha }^{\aleph _{\beta }}=\aleph _{\alpha +1}\,.}$

These were what was to be proved. JRSpriggs (talk) 10:05, 13 November 2012 (UTC)

## Result by Laszlo Patai

Continuum hypothesis#The generalized continuum hypothesis contains the following statement, which is tagged as needing a citation:

"A recent result of Carmi Merimovich shows that, for each n≥1, it is consistent with ZFC that for each κ, 2κ is the nth successor of κ. On the other hand, Laszlo Patai proved, that if γ is an ordinal and for each infinite cardinal κ, 2κ is the γth successor of κ, then γ is finite."

Are those two results published? Using Google Scholar, I have been unable to locate a source by Laszlo Patai containing this statement. The source by Merimovich which the first part is apparently based on might be

{{#invoke:Citation/CS1|citation |CitationClass=journal }}

Is there a source for the part attributed to Laszlo Patai? -- Toshio Yamaguchi 13:07, 11 July 2013 (UTC)

Ladislaus Patai wrote a paper "Über die Reihe der unendlichen Kardinalzahlen" in 1926, published in "Mathematische Zeitschrift" in 1928. But I cannot find this theorem there.
The proof is easy by today's standards:
Assume that ${\displaystyle \gamma \geq \omega }$, so ${\displaystyle 1+\gamma =\gamma }$. Assume ${\displaystyle 2^{\aleph _{\alpha }}=\aleph _{\alpha +\gamma }}$ for all ${\displaystyle \alpha }$. Let ${\displaystyle \delta }$ be minimal such that ${\displaystyle \gamma +\gamma <\gamma +\delta +\gamma }$. Note that ${\displaystyle \delta }$ has to be a limit ordinal ${\displaystyle \leq \gamma }$, and ${\displaystyle \aleph _{\gamma +\delta }}$ must be a singular cardinal.
Why? First, ${\displaystyle \alpha <\delta \Rightarrow \gamma +\gamma =\gamma +\alpha +\gamma =\gamma +\alpha +1+\gamma \Rightarrow \alpha +1<\delta }$.
Second: Clearly ${\displaystyle \delta \leq \gamma }$, as ${\displaystyle \gamma +\gamma <\gamma +\gamma +\gamma }$.
Third: ${\displaystyle cf(\aleph _{\gamma +\delta })=cf(\delta )\leq \delta \leq \gamma <\gamma +\delta \leq \aleph _{\gamma +\delta }}$.
But the sequence ${\displaystyle (2^{\aleph _{\alpha }}:\alpha <\gamma +\delta )}$ is eventually constant with value ${\displaystyle \aleph _{\gamma +\gamma }}$, so by the Bukovsky-Hechler theorem also ${\displaystyle 2^{\aleph _{\gamma +\delta }}}$ has this value ${\displaystyle \aleph _{\gamma +\gamma }<\aleph _{\gamma +\delta +\gamma }}$, contradicting the assumption.
--Aleph4 (talk) 16:44, 13 September 2013 (UTC)

## AC or ~AC

Actually, in ZF, I think the Aleph hypothesis does imply the axiom of choice. It doesn't in ZFU (with urelements) or ZF- (- regularity). Let me see. I'll have to check my references, but

${\displaystyle \mathrm {AH} :(\forall \alpha )2^{\aleph _{\alpha }}=\aleph _{\alpha +1}}$

implies PW:

The power set of a well-ordered set can be well-ordered.

implies AC. — Arthur Rubin (talk) 15:50, 3 September 2013 (UTC)

Yes, indeed.—Emil J. 17:11, 3 September 2013 (UTC)
Found a specific theorem. Theorem 13.3.1 at {{#invoke:citation/CS1|citation

|CitationClass=book }} shows that, under some circumstances, in NBGU, AH implies the well-ordering theorem, which is known to be equivalent to the axiom of choice. Those circumstances are met if there are no urelements, as in NBG set theory. I'll restore the original comment that AH implies AC. — Arthur Rubin (talk) 02:54, 4 September 2013 (UTC)

### ZF + AH -> AC

Obviously, ZF + AC -> ( GCH <-> AH ). And, as I showed at Talk:Axiom of choice/Archive 4#Another try, ZF + GCH -> AC. If one can also show that ZF + AH -> AC, then it follows that ZF -> ( GCH <-> AH ).

It suffices to show for all ordinals α that Vα can be injected into the ordinals. However, there is a difficulty with the argument by Arthur Rubin above because it does not deal with the case that α is a limit ordinal. If one simply tries to aggregate the sequence of injections together, one runs into the fact that one is implicitly using AC to select one injection (or well ordering) at each ordinal below α.

This problem can be fixed by restricting ourselves to just picking one item provided that it is sufficiently large. Choose a single bijection f from P(ωα+1) to ωα+2. We will use this to construct a bijection from Vω+α to an ordinal between ωα+1 and ωα+2.

Suppose S is any subset of ωα+1. Let g(S) = ωα+2·rank(S) + f(S). Let h be the result of collapsing g to squeeze out the holes in its image. Let l be defined inductively on Vω+α by l(T) = h( { l(U) | UT } ). Then l is the desired injection from Vω+α to the ordinals. We use AH again (no use of AC here though) to verify that the elements of rank ω+β are mapped into a range beginning between ωβ and ωβ+1 and ending between ωβ+1 and ωβ+2; which fact is needed to make sure that our original choice of f had a sufficiently large domain. JRSpriggs (talk) 04:53, 6 September 2013 (UTC)

I'm not sure I follow your proof. The proof in Set Theory for the Mathematician that PW implies AC is as follows:
For any fixed α we will show by transfinite recursion on β ≤ α that Vβ can be well-ordered, by constructing a specific relation Sβ (extending Sγ for γ < β)
Let λ = H(Vα) By PW, P(λ) can be well-ordered by a specific relation R
If β is a limit ordinal (or 0), we can construct ${\displaystyle S_{\beta }=\bigcup _{\gamma <\beta }S_{\gamma }}$.
If β = γ+1, then we find the order type of Vγ under Sγ is an ordinal δ, and there is a unique order-preserving function fγ mapping Vγ onto δ. As Vγ is a subset of Vα, δ < λ We then define a function g from Vβ = P(Vγ) into P(λ) by
gγ(x) = the range of fγ restricted to x.
Finally, we define Sβ by:
${\displaystyle xS_{\beta }y=(x\in V_{\gamma }\land y\in V_{\gamma }\land xS_{\gamma }y)\lor (x\in V_{\gamma }\land y\in V_{\beta }-V_{\gamma })\lor (x\in V_{\beta }-V_{\gamma }\land y\in V_{\beta }-V_{\gamma }\land g_{\gamma }(x)Rg_{\gamma }(y))}$
This proof shows that, for any α, Vα can be well-ordered. But any set is a subset of some Vα, so it follows that any set can be well-ordered. — Arthur Rubin (talk) 10:41, 6 September 2013 (UTC)
Never mind. Your proof is the same as mine, as we can show (also by transfinite induction) that λω+α = ωα+1. — Arthur Rubin (talk) 10:54, 6 September 2013 (UTC)
Yes, that is an essentially similar way of proving it. I especially like the use of Hartog's number as a way of avoiding having to figure out what is large enough domain for f. I wish I had thought of that. JRSpriggs (talk) 06:31, 7 September 2013 (UTC)

## AC and the two versions of GCH

User:JRSpriggs made a short remark in a recent edit correcting (thank you!) my previous edit. I would like to expand this remark here for two reasons: I want to help others avoid the mistake that I had made; also, perhaps we can add a short form of this remark into the article itself.

I will write Z0 for ZF without the axiom of foundation and without the axiom of replacement.

There are two versions of GCH:

GCHpower: For all sets ${\displaystyle X}$, every subset ${\displaystyle Y\subseteq P(X)}$ of the power set of ${\displaystyle X}$ either is equinumerous with ${\displaystyle P(X)}$ or can be injected into ${\displaystyle X}$.
GCHaleph: For all ordinals ${\displaystyle \alpha }$ we have ${\displaystyle 2^{\aleph _{\alpha }}=\aleph _{\alpha +1}}$.

So the two versions are indeed equivalent over the base theory ZF. Still I think that the (apparently) "stronger" version GCHpower should be mentioned as the GCH. When we mention Sierpinski's proof, we should clarify that he proved AC from GCHpower.

(In ZF, AC also follows from GCHaleph, but the interesting part is the implication from "P(well-order)=well-orderable" to AC, which has nothing to do with cardinal arithmetic. Was Sierpinski also the author of this theorem?)

--Aleph4 (talk) 12:48, 12 September 2013 (UTC)

You did notice this is discussed in the section above, right?—Emil J. 13:47, 12 September 2013 (UTC)

Yes, I should have referenced the preceding discussion, and adapted my notation. Sorry. The preceding discussion gives the proof of (1→)2→3. My point is that the different character of the two implications between GCH (GCHpower) and AH (GCHaleph) is not made clear in the article. (But if I cannot make this clear on the discussion page, I won't even try on the article page.) --Aleph4 (talk) 15:58, 13 September 2013 (UTC)

## ZF & NGB

The lead has been going back and forth for a few days w/o me being involved except for a revert. Does the lead need to mention NGB set theory? Perhaps so (I have no idea), but in that case, the relationship between ZF and NGB set theories must be described - otherwise it's just confusing. Please discuss it here, it's what the talk page is for. I will not interfere. YohanN7 (talk) 22:27, 5 January 2014 (UTC)

I might have added some more details as Eleuther did were it not for the fact that this is the lead which is merely supposed to summarize the article. The details are already available in Continuum hypothesis#Impossibility of proof and disproof in ZFC at the second paragraph. Godel's proof can easily be adapted to ZF, so there is no need to complicate things by dragging Von Neumann–Bernays–Gödel set theory into it. JRSpriggs (talk) 05:12, 6 January 2014 (UTC)
The key sentence (from Von Neumann–Bernays–Gödel set theory) seems to be A statement in the language of ZFC is provable in NBG if and only if it is provable in ZFC. As far as I can tell, two sentences would be needed in the second paragraph if mention of NGB is to be included here. It could also be inserted as a parenthetical remark - or as a "popup" citation; Gödel used NGB set theory in which statements are provable if and only if they are provable in ZFC - if people feel strongly enough about it. Just a thought. YohanN7 (talk) 13:47, 6 January 2014 (UTC)
The problem with the lead is some clumsy prose, which can be improved, and the incorrect off-hand assertion that ZF is "the standard foundation of modern mathematics," which is simply not the case. Most mathematicians grant that set theory has a foundational role, but I doubt that most would go so far as to select a particular axiomatization as "THE" foundation. NBG is an attractive alternative, and there are many others --- it's an area of active research. However, the lead of this article isn't the place to discuss the issue. My initial idea was just to water down the incorrect assertion to "the standard expression of set theory," acknowledging that ZF is the most popular axiomatization. But that was reverted, as was my most recent attempt to simply remove the incorrect assertion. The article seems to be under the protection of a ZF advocate. Eleuther (talk) 07:24, 7 January 2014 (UTC)
I have to agree partially with Eleuther here, though possibly for different reasons. Calling ZF "the foundation" is a fundamentally formalist POV. Merely claiming that there is a foundation is a foundationalist POV. This is pretty important in context, because if you call ZF "the" foundation, and then talk about CH being neither provable nor refutable from ZF, then you over-weight the position that CH can't be decided in any way at all, which is controversial. --Trovatore (talk) 08:10, 7 January 2014 (UTC)
I see the point as well. I did the revert because, if the CH and ~CH aren't provable from, say, the axioms of field, then it isn't really interesting. What is interesting, in fact it's the main point, is that you can bring on most of all of mathematics based on most all of established axioms of set theory and yet fail to prove CH or ~CH.
This must be communicated to the reader in the lead. It can surely be done in a neutral tone. As of present, the lead uses common POVs as Trovatore points out (the most common ones I suppose). YohanN7 (talk) 13:21, 7 January 2014 (UTC)
Yeah, needs thought. I just tweaked the sentence to reverse the order of "proved" and "disproved" and make it clear that they're parallel (this has no implication as to whether a proof or disproof would be more surprising; I just think "proved nor disproved" sounds better than "disproved nor proved"). I took another look at the ZF thing and couldn't immediately think of a good solution.
One more monkey wrench to throw into the mix: If we're going to talk about a particular formal theory, I think ZFC is a better choice than ZF. It's not even completely clear which formulation is the "real CH" in the absence of choice. --Trovatore (talk) 06:39, 9 January 2014 (UTC)

### "disproved nor proved"

JR's point seems to be that Goedel's contribution is listed first and Cohen's second (same as the chronological order). But the unusual word order "disproved nor proved" really does not get this across; it just looks weird. I suggest that if we want to say that Goedel showed it couldn't be disproved in ZFC and Cohen showed it couldn't be proved in ZFC, then that is what needs to be said. But at that level of the lead, I don't know that we need to get that specific, and I think we should just go back to "proved nor disproved".

Again, it should be ZFC, not ZF. --Trovatore (talk) 07:41, 9 January 2014 (UTC)
Yes, ZFC if anything. Also, could we get rid of the annoyingly prudent "if ZF set theory is consistent" clause in the lead? Pathological cases aren't of interest to the general reader (and hardly anyone else). (How on earth has it not become standard to implicitly assume consistency in the math lingo.) YohanN7 (talk) 15:57, 9 January 2014 (UTC)
Because it might turn out to be inconsistent. Tkuvho (talk) 16:00, 9 January 2014 (UTC)
Yes, and in the intervening 101001000 years we all have to use awkward language in encyclopedias and elsewhere so that we don't forget. You miss the point. In the rest of the article we could be as precise as we wish, provided the sun rises tomorrow. YohanN7 (talk) 16:14, 9 January 2014 (UTC)
You appear to be comfortable in your faith that the sun is less likely to rise tomorrow than ZFC is likely to be consistent, and you are certainly entitled to it. However issues of faith may constitute POV and shouldn't necessarily influence issues of content. Tkuvho (talk) 16:35, 9 January 2014 (UTC)
No, I am not - provided of course that both the sun and ZFC exist. I am just expressing my surprise that mathematics hasn't rationalized its internal language in this case so that it becomes less cumbersome. In most contexts there are obvious implicit assumptions not spelled out each and every time. If a not very obvious assumption is used then it is spelled out. YohanN7 (talk) 18:14, 9 January 2014 (UTC)
JRSpriggs, are you going to address this? I continue to object to "disproved nor proved". The unusual word order gives the reader a problem to solve; the fraction of readers who come up with your intended solution, I expect to be small. There is just no real indication that the word order corresponds to which mathematician did what. I'm concerned that some readers will come up with an unwanted solution, such as losing the parallelism between "disproved" and "proved". --Trovatore (talk) 23:17, 9 January 2014 (UTC)
Would "deduced" be better than "proved"? Tkuvho (talk) 10:10, 13 January 2014 (UTC)

## "a" or "the" standard

There has been some back-and-forth on whether ZF is "a standard foundation" or "the standard foundation". I feel that using the definite article is a POV. Category theory provides another "standard" foundation, for example. There are other varieties of set theory that are similarly considered "standard" by subsets (no pun intended) of the mathematical public, such as MK or NBG. Tkuvho (talk) 08:23, 9 January 2014 (UTC)

Yes, as I explained in the section above, I think that needs to change. The notion that mathematics is founded on formalism at all is itself problematic. (And for the third time, whatever we say about foundational-ness, it should be ZFC, not ZF.)
How about something like "the most common set of formal axioms for set theory, which can be used to derive most of standard mathematics"? --Trovatore (talk) 08:28, 9 January 2014 (UTC)
This would involve explaining that the first occurrence of set is in a metamathematical sense, whereas the second in the technical sense of set theory :-) Also this seems a bit wordy. So long as one uses the indefinite article we stay away from the claim that mathematics is "founded" on formalism. So all in all "a standard foundation" seems appropriate. Tkuvho (talk) 09:07, 9 January 2014 (UTC)
I agree with all of this. The issue is too complicated to phrase briefly, and anyway, it is out of place in the lead of this article. The simplest approach is just to remove the problematic foundational claim, and replace it with something more neutral, like "the most common axiomatization of set theory." I tried that however, and was reverted. I won't try it again, but I will support it if someone else does it. Ideally, JRSpriggs would be the one to make the change. Eleuther (talk) 10:56, 9 January 2014 (UTC)
I think the meaning of "standard" (as opposed to, say, "canonical") is precisely "the most common". I don't think this involves any ontological commitments at all, but is rather an accurate description of current mathematical practice. Tkuvho (talk) 13:29, 9 January 2014 (UTC)
I disagree. The term "standard" has a lot more weight than "most common." We don't describe the Democratic members of the US Senate as standard, and the Republicans as non-standard. "Standard" implies not just a majority, but a majority consensus that the standard should be accepted as standard for some reason. Eleuther (talk) 23:27, 9 January 2014 (UTC)
Agreeing partly with Trovatore, mathematicians just do what they do (informally). Logicians come along later and clean up the mess by formalizing it. So I am not saying that mathematicians refer back to ZFC when they work. Rather what I am saying is that logicians have found that what mathematicians do (in most cases, when they are consistent) can be formalized by the axioms of ZFC. Other foundational systems do not fit so well with what mathematicians actually do. NBG and MK are just slight variations on ZFC. Frankly, I have no intuition for NF (new foundations for set theory) or topos (category theory) so I do not see how they could be the basis of mathematics. JRSpriggs (talk) 09:31, 10 January 2014 (UTC)
To me, the question is not really about alternative foundations; it's about natural strengthenings of ZFC. ZFC does not decide the issue, but some natural strengthening, the addition of some axiom that we can come to the conclusion is the right one to add, might. The natural place to start is large cardinal axioms; it's pretty clear that these axioms (except the ones that turn out to lead to contradictions) are "correct", and their negations are "incorrect". That's why it's misleading to state that we are "free" to add either a statement or its negation to ZFC, if both are consistent; they may both be consistent, but one may be better than the other, in a way similar to the way large-cardinal axioms are better than their negations.
Now, large-cardinal axioms themselves, at least the sort we know about, do not decide CH either, but that does not settle the matter; there might be other axioms for which we can make the judgment that they are better than their negations, and which do decide CH. --Trovatore (talk) 09:59, 10 January 2014 (UTC)

## "the other axioms of set theory"

I have expressed qualified support for Eleuther's concerns, but I do not think this latest effort works very well:

In 1963, Paul Cohen proved that the hypothesis is independent of the other axioms of set theory, based on earlier work by Kurt Gödel in 1940. This surprising result means that one is free to assume that such sets exist or that they do not. Cohen was awarded the Fields Medal in 1966 for his proof.

The most problematic aspect is the second sentence, which is both of unclear meaning and seriously POV if given any substantive meaning. I have removed that sentence.

But I'm not all that thrilled with the rest of it either, though I agree it is simpler. Mainly I think the phrasing "independent of the other axioms of set theory" is problematic. First, "other" axioms of set theory suggests that CH is an axiom of set theory, which is an unusual position. Then too, " 'the' other axioms of set theory" suggests that there's a canonical list of axioms of set theory, which I thought was Eleuther's complaint in the first place. --Trovatore (talk) 00:31, 10 January 2014 (UTC)

If you want input from a non-mathematician (who might still know a little something about math): The lead is not good. It manages to say that CH is independent from the (other) axioms of set theory. The general reader would react with "so what".
• CH is a mathematical statement
• CH was perhaps the most prestigious mathematical problem of the last century
• It cannot be proved using (standard) mathematics
• It cannot be disproved using (standard) mathematics
• Therefore Hilbert's first problem cannot be solved in it's original form
• Is Hilbert's first problem resolved?
• Ongoing work towards resolving CH
The lead is now extremely short. There is plenty of space to beef out a little (on what was the most formidable (yes, POV) mathematical problem of the last century.
The version of a couple of weeks ago was perhaps bad, but this is decidedly worse. Actually, the main message that the general reader will pick up from the lead is that Cohen got a fields medal. YohanN7 (talk) 01:22, 10 January 2014 (UTC)
Actually, I don't think your bullet point about "Hilbert's first problem cannot be solved in its original form" is so clear. Hilbert asked for a proof of CH (the original wording does not seem to have contemplated a disproof, if I remember correctly, but I suppose he would also have accepted that). But he didn't specify any limitations on what axioms might be used, and ZFC had not at the time been formulated, so he could not have meant ZFC. If Woodin's arguments around the year 2000 had eventually been accepted as having established ~CH, would that have satisfied Hilbert? I don't see any way we can know. --Trovatore (talk) 01:58, 10 January 2014 (UTC)
The "bullets" are just bullets, not a draft for a new lead. But surely, there was a notion of "proof" even back in 1900 though we today would probably define "proof" as "proof in ZFC" by default (though it is customary to mention choice if used). I never heard about Hilbert objecting to ZFC (he died 1943) so I don't think the conclusion "Hilbert's first problem cannot be solved in it's original form" is too wrong or too unsupported or uncontroversial. Also, your discussion above could (in modified form) provide some content for bullets #5 and #6. Woodins and others work in the modern era is precisely the thing I mean with the last bullet.
I maintain that the lead must not be totally impeccable technically, or even impeccable POV-wise provided the most common POV is used, if that hinders just about everything interesting to be even mentioned. YohanN7 (talk) 02:56, 10 January 2014 (UTC)
Here's the point: What the independence result says is that you can't decide CH if you're limited to certain axioms. It's certainly interesting that that collection of axioms is sufficient to formalize virtually all of "normal" math, and still doesn't decide CH. But it doesn't follow, by itself, that you can't decide CH in some other way, if you're not limited to those axioms.
So the question is not at all whether Hilbert would have "objected" to ZFC (meaning, presumably, objected to one or more of the axioms), but whether his question was constrained to using only ZFC, and not more. And clearly, given that ZFC had not been formulated, the question could not have been constrained to using exactly ZFC. So you would have to argue that Hilbert was implicitly placing limitations that would constrain his methods to be less-than-or-equal-to ZFC, and while that may be true, I don't see how you establish it.
The lead must not over-weight the position that CH is "absolutely undecidable". I think the best way to avoid it is to call out a set of axioms explicitly, as the old lead did, but explain that that set is enough to do most "normal" math but not decide CH. --Trovatore (talk) 03:10, 10 January 2014 (UTC)
I understand what you mean, and again your answer contains some substance that could go into the article. I agree that the former lead was better, in the sense that it was explicit about at least something. I suggest your version of 00:13 January 10 is moved back for now. YohanN7 (talk) 04:40, 10 January 2014 (UTC)

## restoring ZFC in the lede

The recent removal of the mention of ZFC from the lead is misguided. The question of CH can only be made precise in the context of ZFC (or ZF) and this should be mentioned. I have the impression that only one editor opposes the inclusion of such an explicit mention. It would be nice if the editors expressed themselves on this limited point. Tkuvho (talk) 09:25, 10 January 2014 (UTC)

Well, I don't think I agree with your second sentence (CH is a precise question in the context of informal set theory; you don't have to specify any formalization). But I do agree with your first sentence; without a formalization, CH per se makes sense, but the independence of CH does not, because independence is a negative result ("it cannot be proved"), and to establish such a result, you can't really get started without a precise limitation on the methods of proof. --Trovatore (talk) 10:12, 10 January 2014 (UTC)
Yes, restore ZFC. B t w, did Cohen base his work on that of Gödel? (The current lead says so.) YohanN7 (talk) 10:26, 10 January 2014 (UTC)
Um, in a certain sense. The original version of forcing was a modification of the construction of Goedel's L, which was the (class) model used to establish the consistency of CH. The original version was called ramified forcing, and I don't really know much about it, because it is generally not taught now. Within a few years the "ramified" part was found to be more trouble than it was worth; modern, unramified forcing no longer looks much like the levels of L.
Still, it's the first time I've ever heard anyone take the view that Cohen's proof was an extension of Goedel's; I'd have to think about it, but off the top of my head I don't think that's true in any very interesting way. (In an uninteresting way, pretty much all set theory done at that time was "based" on Goedel's work, because Goedel was the first person to really do serious set-theory-as-set-theory beyond what in retrospect, with no disrespect to the brilliant men who developed them, we can call "elementary methods".) --Trovatore (talk) 10:40, 10 January 2014 (UTC)
Gödel showed that CH is consistent with NBG. That was the first hard part of the Hilbert problem. Cohen then assumed Gödel's result, and incorporated it in his proof of the independence of CH from ZF (ZF being a proper subset of NBG), the second hard part. That's the sense in which Cohen's proof is an extension of Gödel's -- Cohen openly assumed and used Gödel's result. That's also the sense in which I said "based on" in the modified lead. If this is the first time you've heard of it, I think you haven't read the proofs very well. Eleuther (talk) 10:46, 12 January 2014 (UTC)
To be more clear, to show that CH is independent of ZF, one must show that both CH and not-CH are consistent with ZF. Gödel proved the first part in 1940, Cohen the second in 1963. Eleuther (talk) 12:31, 12 January 2014 (UTC)
It did occur to me that you might be using "based on" in that sense, but in that case I don't think it's very good wording. If you have a result that breaks naturally into two pieces, as here, and one researcher proves one piece and then another finishes it off later by proving the other, I think that's the better description. You say Cohen "assumed" Goedel's result, but that's a very odd way of putting it; Cohen's proof stands alone as a proof of Con(ZFC+¬CH). There is no need to "assume" Con(ZFC+CH) in that proof. --Trovatore (talk) 17:43, 12 January 2014 (UTC)
I replaced "based on" by "complementing", hope this is better. Tkuvho (talk) 10:06, 13 January 2014 (UTC)
User:Eleuther has already made three reverts of the material in the introduction. Continuing this behavior may end you up in a block. Tkuvho (talk) 13:10, 12 January 2014 (UTC)

## Multiverse approach to set theory

I may be old-fashioned, but I don't see the specific relevance of Hampkins' paper to CH; although I haven't yet read a the paper, it would seem to apply to any independent (in some sense) statement, not just CH.

Even assuming the paper is considered significant in the field. — Arthur Rubin (talk) 13:51, 27 January 2014 (UTC)

Hamkins has a more general notion of a "switch", namely an axiom that can be adjusted at will to either be satisfied or violated (like the CH) by passing to a larger model. However, his discussion does tend to center on CH, and has immediate relevance to the philosophical debate around CH. Tkuvho (talk) 15:43, 27 January 2014 (UTC)