Compactness
See Notes:Compactness and sequences - I think there's a different definition for metric spaces, I have not seen a proof that the metric one [ilmath]\implies[/ilmath] this one
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]
