Difference between revisions of "Compactness"
From Maths
m |
m (→Lemma for a set being compact) |
||
| Line 19: | Line 19: | ||
That is to say that <math>Y</math> is compact if and only if every covering of <math>Y</math> by sets open in <math>X</math> contains a finite subcovering covering <math>Y</math> | That is to say that <math>Y</math> is compact if and only if every covering of <math>Y</math> by sets open in <math>X</math> contains a finite subcovering covering <math>Y</math> | ||
| − | + | {{Begin Theorem}} | |
| − | + | Theorem: A set {{M|Y\subseteq X}} is a compact space (considered with the subspace topology) of {{M|(X,\mathcal{J})}} ''if and only if'' 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> | + | {{Begin Proof}} |
| + | '''{{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}}) | ||
| − | 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 [[Subspace topology|being a subspace]]) | + | :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 [[Subspace topology|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)''' | + | :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>. | + | :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 <br /> | + | :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 <br /> |
| − | <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><br /> | + | :<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><br /> |
| − | 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><br /> | + | :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><br /> |
| − | then by the [[Implies and subset relation|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> | + | :then by the [[Implies and subset relation|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> |
| − | Lastly, as <math>\mathcal{A}</math> was a covering <math>\cup_{\alpha\in I}A_\alpha=Y</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|implies and subset relation]] we have:<br /> | + | ::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|implies and subset relation]] we have:<br /> |
| − | <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> | + | ::<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> |
| − | 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> | + | ::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> | + | ::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> |
| − | + | ||
| − | + | ||
| − | |||
| − | |||
| − | + | '''{{M|(Y,\mathcal{J}_\text{subspace})}} is compact <math>\impliedby</math> every covering of {{M|Y}} by sets open in {{M|X}} 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> | |
| − | Thus the corresponding finite subcollection of <math>\mathcal{A}'</math> covers <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 {{M|\mathcal{A} }} 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> | ||
| + | {{End Proof}} | ||
| + | {{End Theorem}} | ||
{{Definition|Topology}} | {{Definition|Topology}} | ||
Revision as of 10:37, 8 April 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 if every open cover (often denoted [math]\mathcal{A}[/math]) of [math]X[/math] contains a finite sub-collection that also covers [math]X[/math]
Lemma for a set being compact
Take a set [math]Y\subset X[/math] in a topological space [math](X,\mathcal{J})[/math].
To say [math]Y[/math] is compact is for [math]Y[/math] to be compact when considered as a subspace of [math](X,\mathcal{J})[/math]
That is to say that [math]Y[/math] is compact if and only if every covering of [math]Y[/math] by sets open in [math]X[/math] contains a finite subcovering covering [math]Y[/math]
Theorem: A set [ilmath]Y\subseteq X[/ilmath] is a compact space (considered with the subspace topology) of [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.
[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]
[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]
