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In [[general topology]], a branch of mathematics, a collection ''A'' of subsets of a set ''X'' is said to have the '''finite intersection property''' if the [[intersection (set theory)|intersection]]  over any finite subcollection of ''A'' is nonempty.
This is a preview for the new '''MathML rendering mode''' (with SVG fallback), which is availble in production for registered users.


A '''centered system of sets''' is a collection of sets with the finite intersection property.
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==Definition==
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Let ''X'' be a set with <math>A=\{A_i\}_{i\in I}</math> a family of subsets of ''X''. Then the collection ''A'' has the finite intersection property (fip), if any finite subcollection ''J'' ⊆ ''I'' has non-empty intersection <math>\bigcap_{i\in J} A_i.</math>


==Discussion==
'''MathML'''
Clearly the empty set cannot belong to any collection with the f.i.p.  The condition is trivially satisfied if the intersection over the entire collection is nonempty (in particular, if the collection itself is empty), and it is also trivially satisfied if the collection is nested, meaning that the collection is [[total order|totally ordered]] by inclusion (equivalently, for any finite subcollection, a particular element of the subcollection is contained in all the other elements of the subcollection), e.g. the [[nested sequence of intervals]] (0, 1/''n''). These are not the only possibilities however. For example, if ''X'' = (0, 1) and for each positive integer ''i'', ''X<sub>i</sub>'' is the set of elements of ''X'' having a decimal expansion with digit 0 in the ''i'''th decimal place, then any finite intersection is nonempty (just take 0 in those finitely many places and 1 in the rest), but the intersection of all ''X<sub>i</sub>'' for ''i'' ≥ 1 is empty, since no element of (0, 1) has all zero digits.
:<math forcemathmode="mathml">E=mc^2</math>


The finite intersection property is useful in formulating an alternative definition of [[compact space|compactness]]: a space is compact if and only if every collection of closed sets satisfying the finite intersection property has nonempty intersection itself.<ref>{{planetmathref|id=4181|title=A space is compact iff any family of closed sets having fip has non-empty intersection}}</ref> This formulation of compactness is used in some proofs of [[Tychonoff's theorem]] and the [[uncountable set|uncountability]] of the [[real number]]s (see next section)
<!--'''PNG'''  (currently default in production)
:<math forcemathmode="png">E=mc^2</math>


==Applications==
'''source'''
'''Theorem.''' Let ''X'' be a [[Compact space|compact]] [[Hausdorff space]] that  satisfies the property that no one-point set is open. If ''X'' has more than one point, then ''X'' is uncountable.
:<math forcemathmode="source">E=mc^2</math> -->


'''Proof.''' We will show that if ''U'' ⊆ ''X'' is nonempty and [[Open set|open]], and if ''x'' is a point of ''X'', then there is a [[Neighbourhood (mathematics)|neighbourhood]] ''V'' ⊂ ''U'' whose [[closure (topology)|closure]] doesn’t contain ''x'' (''x'' may or may not be in ''U''). Choose ''y'' in ''U'' different from ''x'' (if ''x'' is in ''U'', then there must exist such a ''y'' for otherwise ''U'' would be an open one point set; if ''x'' isn’t in ''U'', this is possible since ''U'' is nonempty). Then by the Hausdorff condition, choose disjoint neighbourhoods ''W'' and ''K'' of ''x'' and ''y'' respectively. Then ''K''&nbsp;∩&nbsp;''U'' will be a neighbourhood of ''y'' contained in ''U'' whose closure doesn’t contain ''x'' as desired.<br />
<span style="color: red">Follow this [https://en.wikipedia.org/wiki/Special:Preferences#mw-prefsection-rendering link] to change your Math rendering settings.</span> You can also add a [https://en.wikipedia.org/wiki/Special:Preferences#mw-prefsection-rendering-skin Custom CSS] to force the MathML/SVG rendering or select different font families. See [https://www.mediawiki.org/wiki/Extension:Math#CSS_for_the_MathML_with_SVG_fallback_mode these examples].


Now suppose ''f'' : '''N''' → ''X'' is a bijection, and let {''x<sub>i</sub>'' : ''i'' ∈ '''N'''} denote the image of ''f''. Let ''X'' be the first open set and choose a neighbourhood ''U''<sub>1</sub> ⊂ ''X'' whose closure doesn’t contain ''x''<sub>1</sub>. Secondly, choose a neighbourhood ''U''<sub>2</sub> ⊂ ''U''<sub>1</sub> whose closure doesn’t contain ''x''<sub>2</sub>. Continue this process whereby choosing a neighbourhood ''U''<sub>''n''+1</sub> ⊂ ''U<sub>n</sub>'' whose closure doesn’t contain ''x''<sub>''n''+1</sub>. Then the collection {''U<sub>i</sub>'' : ''i'' ∈ '''N'''} satisfies the finite intersection property and hence the intersection of their closures is nonempty (by the compactness of ''X''). Therefore there is a point ''x'' in this intersection. No ''x<sub>i</sub>'' can belong to this intersection because ''x<sub>i</sub>'' doesn’t belong to the closure of ''U<sub>i</sub>''. This means that ''x'' is not equal to ''x<sub>i</sub>'' for all ''i'' and ''f'' is not surjective; a contradiction. Therefore, ''X'' is uncountable.
==Demos==


All the conditions in the statement of the theorem are necessary:
Here are some [https://commons.wikimedia.org/w/index.php?title=Special:ListFiles/Frederic.wang demos]:


1. We cannot eliminate the Hausdorff condition; a countable set with the [[indiscrete topology]] is compact, has more than one point, and satisfies the property that no one point sets are open, but is not uncountable.


2. We cannot eliminate the compactness condition as the set of all rational numbers shows.  
* accessibility:
** Safari + VoiceOver: [https://commons.wikimedia.org/wiki/File:VoiceOver-Mac-Safari.ogv video only], [[File:Voiceover-mathml-example-1.wav|thumb|Voiceover-mathml-example-1]], [[File:Voiceover-mathml-example-2.wav|thumb|Voiceover-mathml-example-2]], [[File:Voiceover-mathml-example-3.wav|thumb|Voiceover-mathml-example-3]], [[File:Voiceover-mathml-example-4.wav|thumb|Voiceover-mathml-example-4]], [[File:Voiceover-mathml-example-5.wav|thumb|Voiceover-mathml-example-5]], [[File:Voiceover-mathml-example-6.wav|thumb|Voiceover-mathml-example-6]], [[File:Voiceover-mathml-example-7.wav|thumb|Voiceover-mathml-example-7]]
** [https://commons.wikimedia.org/wiki/File:MathPlayer-Audio-Windows7-InternetExplorer.ogg Internet Explorer + MathPlayer (audio)]
** [https://commons.wikimedia.org/wiki/File:MathPlayer-SynchronizedHighlighting-WIndows7-InternetExplorer.png Internet Explorer + MathPlayer (synchronized highlighting)]
** [https://commons.wikimedia.org/wiki/File:MathPlayer-Braille-Windows7-InternetExplorer.png Internet Explorer + MathPlayer (braille)]
** NVDA+MathPlayer: [[File:Nvda-mathml-example-1.wav|thumb|Nvda-mathml-example-1]], [[File:Nvda-mathml-example-2.wav|thumb|Nvda-mathml-example-2]], [[File:Nvda-mathml-example-3.wav|thumb|Nvda-mathml-example-3]], [[File:Nvda-mathml-example-4.wav|thumb|Nvda-mathml-example-4]], [[File:Nvda-mathml-example-5.wav|thumb|Nvda-mathml-example-5]], [[File:Nvda-mathml-example-6.wav|thumb|Nvda-mathml-example-6]], [[File:Nvda-mathml-example-7.wav|thumb|Nvda-mathml-example-7]].
** Orca: There is ongoing work, but no support at all at the moment [[File:Orca-mathml-example-1.wav|thumb|Orca-mathml-example-1]], [[File:Orca-mathml-example-2.wav|thumb|Orca-mathml-example-2]], [[File:Orca-mathml-example-3.wav|thumb|Orca-mathml-example-3]], [[File:Orca-mathml-example-4.wav|thumb|Orca-mathml-example-4]], [[File:Orca-mathml-example-5.wav|thumb|Orca-mathml-example-5]], [[File:Orca-mathml-example-6.wav|thumb|Orca-mathml-example-6]], [[File:Orca-mathml-example-7.wav|thumb|Orca-mathml-example-7]].
** From our testing, ChromeVox and JAWS are not able to read the formulas generated by the MathML mode.


3. We cannot eliminate the condition that one point sets cannot be open as a finite space given the [[discrete topology]] shows.
==Test pages ==


'''Corollary.''' Every closed interval [''a'',&nbsp;''b''] with ''a''&nbsp;<&nbsp;''b'' is uncountable. Therefore, '''R''' is uncountable.
To test the '''MathML''', '''PNG''', and '''source''' rendering modes, please go to one of the following test pages:
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*[[Help:Formula]]


'''Corollary.''' Every [[Perfect space|perfect]], [[Locally compact space|locally compact]] Hausdorff space is uncountable.
*[[Inputtypes|Inputtypes (private Wikis only)]]
 
*[[Url2Image|Url2Image (private Wikis only)]]
'''Proof.''' Let ''X'' be a perfect, compact, Hausdorff space, then the theorem immediately implies that ''X'' is uncountable. If ''X'' is a perfect, locally compact Hausdorff space which is not compact, then the [[one-point compactification]] of ''X'' is a perfect, compact Hausdorff space. Therefore the one point compactification of ''X'' is uncountable. Since removing a point from an uncountable set still leaves an uncountable set, ''X'' is uncountable as well.
==Bug reporting==
 
If you find any bugs, please report them at [https://bugzilla.wikimedia.org/enter_bug.cgi?product=MediaWiki%20extensions&component=Math&version=master&short_desc=Math-preview%20rendering%20problem Bugzilla], or write an email to math_bugs (at) ckurs (dot) de .
==Examples==
A [[filter (topology)|filter]] has the finite intersection property by definition.
 
== Theorems ==
Let ''X'' be nonempty, ''F'' ⊆ 2<sup>''X''</sup>, ''F'' having the finite intersection property. Then there exists an ''F''′ [[ultrafilter]] (in 2<sup>''X''</sup>) such that ''F'' ⊆ ''F''′.
 
See details and proof in {{harvtxt|Csirmaz|Hajnal|1994}}.<ref>{{citation|last1=Csirmaz|first1=László|last2=Hajnal|first2=András|author2-link=András Hajnal|title=Matematikai logika|publisher=[[Eötvös Loránd University]]|location=Budapest|year=1994|url=http://www.renyi.hu/~csirmaz/|format=In Hungarian}}.</ref> This result is known as [[ultrafilter lemma]].
 
==Variants==
A family of sets ''A'' has the '''strong finite intersection property''' (sfip), if every finite subfamily of ''A'' has infinite intersection.
 
== References ==
<references/>
* {{planetmathref|id=4178|title=Finite intersection property}}
 
{{DEFAULTSORT:Finite Intersection Property}}
[[Category:General topology]]
[[Category:Set families]]

Latest revision as of 23:52, 15 September 2019

This is a preview for the new MathML rendering mode (with SVG fallback), which is availble in production for registered users.

If you would like use the MathML rendering mode, you need a wikipedia user account that can be registered here [[1]]

  • Only registered users will be able to execute this rendering mode.
  • Note: you need not enter a email address (nor any other private information). Please do not use a password that you use elsewhere.

Registered users will be able to choose between the following three rendering modes:

MathML


Follow this link to change your Math rendering settings. You can also add a Custom CSS to force the MathML/SVG rendering or select different font families. See these examples.

Demos

Here are some demos:


Test pages

To test the MathML, PNG, and source rendering modes, please go to one of the following test pages:

Bug reporting

If you find any bugs, please report them at Bugzilla, or write an email to math_bugs (at) ckurs (dot) de .