Difference between revisions of "Compactness"
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==Definition== | ==Definition== | ||
| − | A [[Topological space|topological space]] is compact if every [[Covering|open cover]] | + | *A [[Topological space|topological space]] is compact<ref name="Topology">Topology - James R. Munkres - Second Edition</ref> if every [[Covering|open cover]] of <math>X</math> contains a finite sub-covering that also covers <math>X</math>. |
| + | That is to say that given an arbitrary collection of sets: | ||
| + | * {{M|1=\mathcal{A}=\{A_\alpha\}_{\alpha\in I} }} such that each {{M|A_\alpha}} is [[Open set|open]] in {{M|X}} and | ||
| + | * {{MM|1=X=\bigcup_{\alpha\in I}A_\alpha}}<ref group="Note">Note that we actually have {{M|X\subseteq\bigcup_{\alpha\in I}A_\alpha}} but as topologies are closed under arbitrary union and contain the set the open sets are subsets of we cannot "exceed {{M|X}}", so we must have {{M|1=X=\bigcup_{\alpha\in I}A_\alpha}}</ref> | ||
| + | The following is true: | ||
| + | * {{M|1=\exists \{i_1,\cdots,i_n\}\subset I}} such that {{MM|1=X=\bigcup_{\alpha\in\{i_1,\cdots,i_n\} }A_\alpha}} | ||
| + | Then {{M|X}} is ''compact''<ref name="Topology"/> | ||
==Lemma for a set being compact== | ==Lemma for a set being compact== | ||
| − | Take a set <math>Y\subset X</math> in a [[Topological space|topological space]] <math>(X,\mathcal{J})</math>. | + | Take a set <math>Y\subset X</math> in a [[Topological space|topological space]] <math>(X,\mathcal{J})</math>. Then to say: |
| − | + | * <math>Y</math> is compact | |
| − | + | Means <math>Y</math> satisfies the definition of compactness when considered as a [[Subspace topology|subspace]] of <math>(X,\mathcal{J})</math> | |
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| − | + | ||
| − | + | ||
{{Begin Theorem}} | {{Begin Theorem}} | ||
| − | Theorem: A set {{M|Y\subseteq X}} is a compact | + | Theorem: A set {{M|Y\subseteq X}} is a compact in {{M|(X,\mathcal{J})}} ''if and only if'' every covering of {{M|Y}} by sets open in {{M|X}} contains a finite subcovering. |
{{Begin Proof}} | {{Begin Proof}} | ||
| + | {{Todo|Redo this proof - it is not very well written}} | ||
'''{{M|(Y,\mathcal{J}_\text{subspace})}} is compact {{M|\implies}} every covering of {{M|Y}} by sets open in {{M|X}} contains a finite subcovering''' | '''{{M|(Y,\mathcal{J}_\text{subspace})}} is compact {{M|\implies}} every covering of {{M|Y}} by sets open in {{M|X}} contains a finite subcovering''' | ||
:Suppose that the space <math>(Y,\mathcal{J}_\text{subspace})</math> is compact and that <math>\mathcal{A}=\{A_\alpha\}_{\alpha\in I}</math> (where each <math>A_\alpha\in\mathcal{J}</math> - that is each set is open in <math>X</math>) is an open covering (which is to say {{M|Y\subseteq\cup_{\alpha\in I}A_\alpha}}) | :Suppose that the space <math>(Y,\mathcal{J}_\text{subspace})</math> is compact and that <math>\mathcal{A}=\{A_\alpha\}_{\alpha\in I}</math> (where each <math>A_\alpha\in\mathcal{J}</math> - that is each set is open in <math>X</math>) is an open covering (which is to say {{M|Y\subseteq\cup_{\alpha\in I}A_\alpha}}) | ||
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{{End Proof}} | {{End Proof}} | ||
{{End Theorem}} | {{End Theorem}} | ||
| + | |||
| + | ==See also== | ||
| + | * [[Subspace topology]] | ||
| + | |||
| + | ==Notes== | ||
| + | <references group="Note"/> | ||
| + | |||
| + | ==References== | ||
| + | <references/> | ||
{{Definition|Topology}} | {{Definition|Topology}} | ||
Revision as of 03:48, 22 June 2015
Not to be confused with Sequential compactness
There are two views here.
- Compactness is a topological property and we cannot say a set is compact, we say it is compact and implicitly consider it with the subspace topology
- We can say "sure that set is compact".
The difference comes into play when we cover a set (take the interval [ilmath][0,5]\subset\mathbb{R} [/ilmath]) with open sets. Suppose we have the covering [ilmath]\{(-1,3),(2,6)\} [/ilmath] this is already finite and covers the interval. The corresponding sets in the subspace topology are [ilmath]\{[0,3),(2,5]\} [/ilmath] which are both open in the subspace topology.
Definition
- A topological space is compact[1] if every open cover of [math]X[/math] contains a finite sub-covering that also covers [math]X[/math].
That is to say that given an arbitrary collection of sets:
- [ilmath]\mathcal{A}=\{A_\alpha\}_{\alpha\in I}[/ilmath] such that each [ilmath]A_\alpha[/ilmath] is open in [ilmath]X[/ilmath] and
- [math]X=\bigcup_{\alpha\in I}A_\alpha[/math][Note 1]
The following is true:
- [ilmath]\exists \{i_1,\cdots,i_n\}\subset I[/ilmath] such that [math]X=\bigcup_{\alpha\in\{i_1,\cdots,i_n\} }A_\alpha[/math]
Then [ilmath]X[/ilmath] is compact[1]
Lemma for a set being compact
Take a set [math]Y\subset X[/math] in a topological space [math](X,\mathcal{J})[/math]. Then to say:
- [math]Y[/math] is compact
Means [math]Y[/math] satisfies the definition of compactness when considered as a subspace of [math](X,\mathcal{J})[/math]
Theorem: A set [ilmath]Y\subseteq X[/ilmath] is a compact in [ilmath](X,\mathcal{J})[/ilmath] if and only if every covering of [ilmath]Y[/ilmath] by sets open in [ilmath]X[/ilmath] contains a finite subcovering.
TODO: Redo this proof - it is not very well written
[ilmath](Y,\mathcal{J}_\text{subspace})[/ilmath] is compact [ilmath]\implies[/ilmath] every covering of [ilmath]Y[/ilmath] by sets open in [ilmath]X[/ilmath] contains a finite subcovering
- Suppose that the space [math](Y,\mathcal{J}_\text{subspace})[/math] is compact and that [math]\mathcal{A}=\{A_\alpha\}_{\alpha\in I}[/math] (where each [math]A_\alpha\in\mathcal{J}[/math] - that is each set is open in [math]X[/math]) is an open covering (which is to say [ilmath]Y\subseteq\cup_{\alpha\in I}A_\alpha[/ilmath])
- Then the collection [math]\{A_\alpha\cap Y|\alpha\in I\}[/math] is a covering of [math]Y[/math] by sets open in [math]Y[/math] (by definition of being a subspace)
- By hypothesis [math]Y[/math] is compact, hence a finite sub-collection [math]\{A_{\alpha_i}\cap Y\}^n_{i=1}[/math] covers [math]Y[/math] (as to be compact every open cover must have a finite subcover)
- Then [math]\{A_{\alpha_i}\}^n_{i=1}[/math] is a sub-collection of [math]\mathcal{A}[/math] that covers [math]Y[/math].
Proof of details
- As The intersection of sets is a subset of each set and [math]\cup^n_{i=1}(A_{\alpha_i}\cap Y)=Y[/math] we see
- [math]x\in\cup^n_{i=1}(A_{\alpha_i}\cap Y)\implies\exists k\in\mathbb{N}\text{ with }1\le k\le n:x\in A_{\alpha_k}\cap Y[/math] [math]\implies x\in A_{\alpha_k}\implies x\in\cup^n_{i=1}A_{\alpha_i}[/math]
- The important part being [math]x\in\cup^n_{i=1}(A_{\alpha_i}\cap Y)\implies x\in\cup^n_{i=1}A_{\alpha_i}[/math]
- then by the implies and subset relation we have [math]Y=\cup^n_{i=1}(A_{\alpha_i}\cap Y)\subset\cup^n_{i=1}A_{\alpha_i}[/math] and conclude [math]Y\subset\cup^n_{i=1}A_{\alpha_i}[/math]
- Warning: this next bit looks funny - do not count on!
- Lastly, as [math]\mathcal{A}[/math] was a covering [math]\cup_{\alpha\in I}A_\alpha=Y[/math].
- It is clear that [math]x\in\cup^n_{i=1}A_{\alpha_i}\implies x\in\cup_{\alpha\in I}A_\alpha[/math] so again implies and subset relation we have:
- [math]\cup^n_{i=1}A_{\alpha_i}\subset\cup_{\alpha\in I}A_\alpha=Y[/math] thus concluding [math]\cup^n_{i=1}A_{\alpha_i}\subset Y[/math]
- It is clear that [math]x\in\cup^n_{i=1}A_{\alpha_i}\implies x\in\cup_{\alpha\in I}A_\alpha[/math] so again implies and subset relation we have:
- Combining [math]Y\subset\cup^n_{i=1}A_{\alpha_i}[/math] and [math]\cup^n_{i=1}A_{\alpha_i}\subset Y[/math] we see [math]\cup^n_{i=1}A_{\alpha_i}=Y[/math]
- Thus [math]\{A_{\alpha_i}\}^n_{i=1}[/math] is a finite covering of [math]Y[/math] consisting of open sets from [math]X[/math]
- End of warning - I've left this here because I must have put it in for a reason!
TODO: What was I hoping to do here?
[ilmath](Y,\mathcal{J}_\text{subspace})[/ilmath] is compact [math]\impliedby[/math] every covering of [ilmath]Y[/ilmath] by sets open in [ilmath]X[/ilmath] contains a finite subcovering
- Suppose that every covering of [math]Y[/math] by sets open in [math]X[/math] contains a finite subcollection covering [math]Y[/math]. We need to show [math]Y[/math] is compact.
- Suppose we have a covering, [math]\mathcal{A}'=\{A'_\alpha\}_{\alpha\in I}[/math] of [math]Y[/math] by sets open in [math]Y[/math]
- For each [math]\alpha[/math] choose an open set [math]A_\alpha[/math] open in [math]X[/math] such that: [math]A'_\alpha=A_\alpha\cap Y[/math]
- Then the collection [math]\mathcal{A}=\{A_\alpha\}_{\alpha\in I}[/math] covers [math]Y[/math]
- By hypothesis we have a finite sub-collection from [ilmath]\mathcal{A} [/ilmath] of things open in [math]X[/math] that cover [math]Y[/math]
- Thus the corresponding finite subcollection of [math]\mathcal{A}'[/math] covers [math]Y[/math]
See also
Notes
- ↑ Note that we actually have [ilmath]X\subseteq\bigcup_{\alpha\in I}A_\alpha[/ilmath] but as topologies are closed under arbitrary union and contain the set the open sets are subsets of we cannot "exceed [ilmath]X[/ilmath]", so we must have [ilmath]X=\bigcup_{\alpha\in I}A_\alpha[/ilmath]
