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. $\square$

  • $\begingroup$ If $S$ is "a cover of $F$ which has no finite subcover", then the elements of $S$ are subsets of $X$. Later on, you write $x\in S$ which does not make any sense because $x$ is an element of $X$, not a subset of it. $\endgroup$ – Henning Makholm Dec 25 '18 at 16:17
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    $\begingroup$ As a more high-level comment, using an indirect proof here is a definite detour. You can prove that $F$ is compact directly, by letting someone give you an open cover of $F$ and showing how to find a finite subcover of it, using the compactness of $K$. $\endgroup$ – Henning Makholm Dec 25 '18 at 16:19
  • $\begingroup$ I see. I believe Rudin's proof is something along that line but I always try to come up with an alternative proof and this was the first proof that came naturally to me. Regarding $x$ not being an element of $S$, is the new edit enough to fix the problem? $\endgroup$ – Sei Sakata Dec 25 '18 at 16:25
  • $\begingroup$ That is a very good point if $M={X}\setminus F$, then $M\cup S$ is also a cover of $K$ $\endgroup$ – Sei Sakata Dec 25 '18 at 16:46

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.

  • $\begingroup$ I believe the reason I defied $G$ to be finite was that I was trying to lay out the definitions before tackling the problem and I failed to realise that it was a redundant condition. But I think in this case I'm not sure if I have to use the property that $K$ is compact. Because what I essentially wanted to show was that if $F$ is not compact, then $K$ cannot be compact either which contradicts the assumption that $K$ was compact. $\endgroup$ – Sei Sakata Dec 25 '18 at 16:58
  • $\begingroup$ This is a fantastic answer. It dives right in to the student's work and details all the misplaced assumptions and confused arguments, and explains how to fix them. It's a lot more work to write this up than to just give a corrected version of the argument. Nicely done. $\endgroup$ – JonathanZ Dec 25 '18 at 18:09
  • $\begingroup$ This link math.stackexchange.com/questions/262828/… addresses the confusion with surrounding the difference between proof by contradiction and contrapositive with some very good answers. $\endgroup$ – Sei Sakata Dec 25 '18 at 20:02

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

Here a way that is not contrapositive.

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.

  • $\begingroup$ This is verbatim the proof given by Rudin(although more readable) and I DO indeed like its simplicity but I believe its always useful to come up with an original proof without first having to look at the "perfect example". $\endgroup$ – Sei Sakata Dec 25 '18 at 16:33
  • $\begingroup$ That can never harm, and it is indeed useful on its own to try things out yourself. So I share in that belief :-). $\endgroup$ – drhab Dec 25 '18 at 16:38

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. $\square$

  • $\begingroup$ this is fine -- however i would argue little is gained in re-writing this. its very similar to rudin's proof in terms of using many of the same ideas/relationships. but writing a proof is still an exercise in itself i suppose! $\endgroup$ – H_1317 Dec 27 '18 at 1:44

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