Compactness/Uniting covers proof

Theorem statement

• A subset $Y\subseteq X$ of a topological space $(X,\mathcal{J})$ is compact (when $Y$ is imbued with the subspace topology)

if and only if

• Every cover by sets open in $X$ has a finite subcover.

This is very important in uniting the two definitions of "open cover" used with compactness. These are:

1. A covering is a collection of sets whose union contains $Y$
   • That is a collection $\{A_\alpha\}_{\alpha\in I}$ where $Y\subseteq\cup_{\alpha\in I}A_\alpha$
2. A covering is a collection of sets whose union is exactly $Y$, by sets open in $Y$ when $Y$ has the subspace topology.
   • That is a collection $\{A_\alpha\}_{\alpha\in I}$ where $Y=\cup_{\alpha\in I}A_\alpha$

Lastly recall:

• A subcovering is a collection of elements of a cover whose union also cover $Y$

I use the definition of "contains" because to cover something is to overlap it, like cover a dog with a blanket, or cover a face with a pillow.

Proof

Suppose that $(Y,\mathcal{J}_\text{subspace})$ is compact $\implies$ every covering consisting of open sets of $(X,\mathcal{J})$ contains a finite subcover.

Let $\{A_\alpha\}_{\alpha\in I}\subseteq\mathcal{J}$ be a family of open sets in $X$ with $Y\subseteq\cup_{\alpha\in I}A_\alpha$

Take $B_\alpha=A_\alpha\cap Y$, then $\{B_\alpha\}_{\alpha\in I}$ is an open (in $Y$) covering of $Y$, that is $Y\subseteq\cup_{\alpha\in I}B_\alpha$ (infact we have $Y=\cup_{\alpha\in I}B_\alpha$)

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Proof of $Y\subseteq\cup_{\alpha\in I}B_\alpha$ (we actually have $Y=\cup_{\alpha\in I}B_\alpha$)

We wish to show that $Y\subseteq\cup_{\alpha\in I}A_\alpha\implies Y\subseteq\cup_{\alpha\in I}(A_\alpha\cap Y)$, using the Implies-subset relation we actually just want to show that:

• Given $Y\subseteq\cup_{\alpha\in I}A_\alpha$ that $y\in Y\implies y\in\cup_{\alpha\in I}(A_\alpha\cap Y)$ - which is what we'll do.
• Note additionally that $y\in\cup_{\alpha\in I}(A_i\cap Y)\iff \exists\beta\in I[y\in A_\beta\wedge y\in Y]$

Let $y\in Y$, then by hypothesis $y\in\cup_{\alpha\in I}A_\alpha\iff\exists \beta\in I[y\in A_\beta]$

It is easily seen that $y\in Y\wedge\exists\beta\in I[y\in A_\beta]\implies\exists\gamma\in I[y\in A_\gamma\wedge y\in Y]$ simply by choosing $\gamma:=\beta$.

Lastly, note that $\exists\gamma\in I[y\in A_\gamma\wedge y\in Y]\iff y\in\cup_{\alpha\in I}(A_\alpha\cap Y)$

• We have shown that $y\in Y\implies y\in\cup_{\alpha\in I}(A_\alpha\cap Y)$ and by the Implies-subset relation we see
• $Y\subseteq\cup_{\alpha\in I}(A_\alpha\cap Y)$ - as required.

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I earlier claimed that actually $Y=\cup_{\alpha\in I}(A_\alpha\cap Y)$ - this isn't important to the proof but it shows something else.

This shows that considering an open covering as a union of sets open in $Y$ whose union is exactly $Y$ is the same as a covering by open sets in $X$ whose union contains (but need not be exactly equal to) $Y$. So we have shown so far that:

• Compact in the subspace with equality for an open covering $\implies$ compact with the open cover of sets in $X$ whose union contains $Y$

Claim: $\cup_{\alpha\in I}(A_\alpha\cap Y)\subseteq Y$

Let $y\in\cup_{\alpha\in I}(A_\alpha\cap Y)$ then:

• $\exists\beta\in I[y\in A_\beta\wedge y\in Y]\iff \exists\beta\in I[y\in(A_\beta\cap Y)]$

As the intersection of sets is a subset of each set we see that (it's trivial to show without this result too, but this uses a general result)

• $(A_\beta\cap Y)\subseteq Y)$ by the implies-subset relation we see immediately that:

$y\in(A_\beta\cap Y)\implies y\in Y$

Thus we have shown that $y\in\cup_{\alpha\in I}(A_\alpha\cap Y)\implies y\in Y$ and finally this means:

• $\cup_{\alpha\in I}(A_\alpha\cap Y)\subseteq Y$

Combining this with $Y\subseteq \cup_{\alpha\in I}(A_\alpha\cap Y)$ above we see that:

• $Y=\cup_{\alpha\in I}(A_\alpha\cap Y)$

This completes the proof

By hypothesis, $Y$ is compact, this means that $\{B_\alpha\}_{\alpha\in I}$ contains a finite subcover

• call this subcover $\{B'_i\}_{i=1}^n$ where each $B'_i\in\{B_\alpha\}_{\alpha\in I}$, now we have $Y\subseteq\cup_{i=1}^n B'_i$ (we actually have equality, see the blue box in the yellow note box above)

As each $B'_i=A'_i\cap Y$ (where $A'_i$ is the corresponding $A_\alpha$ for the $B_\alpha$ that $B'_i$ represents) we see that $\{A_i\}_{i=1}^n$ is a finite subcover by sets open in $X$

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Proof of: $Y\subseteq\cup_{i=1}^nB'_i\implies$ $Y\subseteq\cup_{i=1}^nA'_i$ (proving that $\{A'_i\}_{i=1}^n$ is an open cover)

For each $i$ we have $B'_i:=A'_i\cap Y$, by invoking the intersection of sets is a subset of each set we note that:

• $B'_i\subseteq A'_i$

We now invoke Union of subsets is a subset of the union

This theorem states that given two families of sets, $\{A_\alpha\}_{\alpha\in I}$ and $\{B_\alpha\}_{\alpha\in I}$ with $\forall\alpha\in I[B_\alpha\subseteq A_\alpha]$ we have $\cup_{\alpha\in I}B_\alpha\subseteq\cup_{\alpha\in I}A_\alpha$

It follows that $Y\subseteq\cup_{i=1}^nB'_i\subseteq\cup_{i=1}^nA'_i$, in particular:

• $Y\subseteq\cup_{i=1}^nA'_i$

This confirms that $\{A'_i\}_{i=1}$ is an open cover by sets in $X$

This completes this half of the proof.

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

Notes

This theorem shows that the two are the same, that is:

• A set $Y\subseteq X$ is compact if every open cover of $Y$ by sets open in $X$ has a finite subcover (that is $Y\subset\text{the covering}$)

and

• A set $Y\subseteq X$ is compact if, when given the Subspace topology, every covering of $Y$ by sets open in $Y$ has a finite subcover.

Are the same, (an if and only if relationship)

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TODO: Make the second part of the proof a bit more hand holdy, it's valid but some of the set theoretic stuff could use proofs

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TODO: References for the definitions and stuff - they are CERTAINLY valid

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