Compactness

Not to be confused with Sequential compactness

There are two views here.

1. 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
2. We can say "sure that set is compact".

The difference comes into play when we cover a set (take the interval $[0,5]\subset\mathbb{R}$) with open sets. Suppose we have the covering $\{(-1,3),(2,6)\}$ this is already finite and covers the interval. The corresponding sets in the subspace topology are $\{[0,3),(2,5]\}$ which are both open in the subspace topology.

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Definition

• A topological space is compact if every open cover of $$X$$ contains a finite sub-covering that also covers $$X$$ .

That is to say that given an arbitrary collection of sets:

• $\mathcal{A}=\{A_\alpha\}_{\alpha\in I}$ such that each $A_\alpha$ is open in $X$ and
• $$X=\bigcup_{\alpha\in I}A_\alpha$$

The following is true:

• $\exists \{i_1,\cdots,i_n\}\subset I$ such that $$X=\bigcup_{\alpha\in\{i_1,\cdots,i_n\} }A_\alpha$$

Then $X$ is compact

Lemma for a set being compact

Take a set

$$Y\subset X$$ in a topological space $$(X,\mathcal{J})$$ . Then to say:

• $$Y$$ is compact

Means

$$Y$$ satisfies the definition of compactness when considered as a subspace of $$(X,\mathcal{J})$$

[Expand]

Theorem: A set $Y\subseteq X$ is a compact in $(X,\mathcal{J})$ if and only if every covering of $Y$ by sets open in $X$ contains a finite subcovering.

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TODO: Redo this proof - it is not very well written

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$(Y,\mathcal{J}_\text{subspace})$ is compact $\implies$ every covering of $Y$ by sets open in $X$ contains a finite subcovering

Suppose that the space $$(Y,\mathcal{J}_\text{subspace})$$ is compact and that $$\mathcal{A}=\{A_\alpha\}_{\alpha\in I}$$ (where each $$A_\alpha\in\mathcal{J}$$ - that is each set is open in $$X$$ ) is an open covering (which is to say $Y\subseteq\cup_{\alpha\in I}A_\alpha$)

Then the collection $$\{A_\alpha\cap Y|\alpha\in I\}$$ is a covering of $$Y$$ by sets open in $$Y$$ (by definition of being a subspace)

By hypothesis $$Y$$ is compact, hence a finite sub-collection $$\{A_{\alpha_i}\cap Y\}^n_{i=1}$$ covers $$Y$$ (as to be compact every open cover must have a finite subcover)

Then $$\{A_{\alpha_i}\}^n_{i=1}$$ is a sub-collection of $$\mathcal{A}$$ that covers $$Y$$ .

Proof of details

As The intersection of sets is a subset of each set and $$\cup^n_{i=1}(A_{\alpha_i}\cap Y)=Y$$ we see

$$x\in\cup^n_{i=1}(A_{\alpha_i}\cap Y)\implies\exists k\in\mathbb{N}\text{ with }1\le k\le n:x\in A_{\alpha_k}\cap Y$$ $$\implies x\in A_{\alpha_k}\implies x\in\cup^n_{i=1}A_{\alpha_i}$$

The important part being $$x\in\cup^n_{i=1}(A_{\alpha_i}\cap Y)\implies x\in\cup^n_{i=1}A_{\alpha_i}$$

then by the implies and subset relation we have $$Y=\cup^n_{i=1}(A_{\alpha_i}\cap Y)\subset\cup^n_{i=1}A_{\alpha_i}$$ and conclude $$Y\subset\cup^n_{i=1}A_{\alpha_i}$$

Warning: this next bit looks funny - do not count on!

Lastly, as $$\mathcal{A}$$ was a covering $$\cup_{\alpha\in I}A_\alpha=Y$$ .

It is clear that $$x\in\cup^n_{i=1}A_{\alpha_i}\implies x\in\cup_{\alpha\in I}A_\alpha$$ so again implies and subset relation we have:

$$\cup^n_{i=1}A_{\alpha_i}\subset\cup_{\alpha\in I}A_\alpha=Y$$ thus concluding $$\cup^n_{i=1}A_{\alpha_i}\subset Y$$

Combining $$Y\subset\cup^n_{i=1}A_{\alpha_i}$$ and $$\cup^n_{i=1}A_{\alpha_i}\subset Y$$ we see $$\cup^n_{i=1}A_{\alpha_i}=Y$$

Thus $$\{A_{\alpha_i}\}^n_{i=1}$$ is a finite covering of $$Y$$ consisting of open sets from $$X$$

End of warning - I've left this here because I must have put it in for a reason!

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TODO: What was I hoping to do here?

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$(Y,\mathcal{J}_\text{subspace})$ is compact $$\impliedby$$ every covering of $Y$ by sets open in $X$ contains a finite subcovering

Suppose that every covering of $$Y$$ by sets open in $$X$$ contains a finite subcollection covering $$Y$$ . We need to show $$Y$$ is compact.

Suppose we have a covering, $$\mathcal{A}'=\{A'_\alpha\}_{\alpha\in I}$$ of $$Y$$ by sets open in $$Y$$

For each $$\alpha$$ choose an open set $$A_\alpha$$ open in $$X$$ such that: $$A'_\alpha=A_\alpha\cap Y$$

Then the collection $$\mathcal{A}=\{A_\alpha\}_{\alpha\in I}$$ covers $$Y$$

By hypothesis we have a finite sub-collection from $\mathcal{A}$ of things open in $$X$$ that cover $$Y$$

Thus the corresponding finite subcollection of $$\mathcal{A}'$$ covers $$Y$$

See also

• Subspace topology