Difference between revisions of "Compactness"

$(Y,\mathcal{J}_\text{subspace})$ 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!

TODO: What was I hoping to do here?

$(Y,\mathcal{J}_\text{subspace})$ 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 $$Y$$