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 $\implies$ this one

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 $$(X,\mathcal{J})$$

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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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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 $$Y$$

See also

• Subspace topology