Closed subsets of compact sets are compact (original proof)

Question

Baby Rudin Theorem 2.35: Suppose $F \subset K \subset X$ where $X$ a metric space. Suppose $F$ closed relative to $K$ and $K$ compact. Then $F$ is compact.

This is my original attempt to prove this theorem.

Here, Rudin's definitions are :

• $p$ is a limit point of a subset of a metric space iff every neighborhood(defined in terms of open balls) of $p$ contains a point in the subset which is not equal to $p$

• Subsets of metric spaces are closed iff it contains all its limit points.

• A point $p$ is an interior point of a subset of a metric space if there exists a neighborhood $N(p)$ of $p$ such that $N(p)$ is contained within the subset.

• Subsets of metric spaces are open iff all its points are interior points.

• Subsets of metric spaces are compact iff every open cover has a finite subcover.

Proof: Suppose $K$ compact and $F \subset K$ closed but not compact. Then there is a cover of $F$ which has no finite subcover say $S$. Let $G=\bigcup^{n}_{i=1}G_i$ be an open cover of $K$ which is finite. Define $M_i=G_i\cap F^c $ where $F^c$ complement of $F$. Then each $M_i$ is open since $G_i$ and $F^c$ are open. Now for any $x \in K$, either $x \in F$ or $x \in F^c$. If $x\in F$, then $x \in S_\alpha $ where $S_\alpha$ is an element of $S$. If $x \in F^c$ then $x \in G_i\cap F^c $ for some $i \in \mathbb{N}$ since $K \subset G$. So if $M=\bigcup^{n}_{i=1} M_i$, then $M \cup S$ is a cover of $K$. But then clearly, this cover has no finite subcover hence $K$ is not compact. $\square$

Answer 1

Your proof is hard to read, has deficiencies (see comment of Henning) and is needlessly complicated.

Here a way that is not contrapositive.

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Let $\mathcal G$ be a collection of open sets that cover $F$.

Then $\mathcal G\cup\{F^{\complement}\}$ is a collection of open sets that covers $K$.

Since $K$ is compact there is a finite subcollection $\{G_1,\dots,G_n\}\cup\{F^{\complement}\}$.

From $F\subseteq K=\left(\bigcup_{i=1}^nG_i\right)\cup F^{\complement}$ it follows that $F\subseteq\bigcup_{i=1}^nG_i$ so apparantly $\{G_1,\dots,G_n\}$ is a finite subcollection of $\mathcal G$ that covers $F$.

Proved is now that every collection of open sets that cover $F$ has a finite subcollection that covers $F$.

That comes to the same as: $F$ is compact.

Answer 2

You're phrasing your proof as:

Suppose $K$ is compact, bla bla bla bla hence $K$ is not compact.

This is the structure of a proof by contradiction, but your underlying idea is not really a proof by contradiction. In particular you're never using the assumption that $K$ is compact anywhere in the "bla bla bla bla" part. So what you really have is an argument that

Assume $F$ is not compact. Then $K$ is not compact.

This is the contrapositive of the implication you wanted to show. Since every implication is logically equivalent to its contrapositive, this gives rise to a formally valid proof -- but pretending it is a proof-by-contradiction by tacking on a never-used assumption that $K$ is compact just makes it confusing and hard to follow. The reader keeps waiting for a place where the assumption is used, but it never comes.

Another point where your proof is needlessly complex is when you say "Let $G$ be an open cover of $K$ which is finite". There's no reason for you to parameterize your proof over an arbitrary open cover when instead you can just decide on a particular finite open cover of $K$ to use, namely setting $G=\{X\}$. Doing so would allow you to avoid a level of indices in the entire rest of the proof.

Actually I suspect these two points are connected: You may have thought that you need to assume that $K$ is compact before you can assert that it has a finite open cover. But that's not what compactness means -- everything has a finite open cover, namely $\{X\}$. Compactness means that the set has finite open covers with special properties (namely being subcovers of an infinite open cover you have thought up somehow), but you don't actually need those special properties in the argument you present.

Answer 3

Refined proof:

Suppose $F$ not compact. Then there is a cover of $F$ which has no finite subcover say $S$. Define $M=\{X\}\setminus F$. Then $M \cup S$ is a cover of $K$. But then clearly, this cover has no finite subcover hence $K$ is not compact. $\square$