# Begging the question in Rudin?

I read this in Theorem 2.35 of Baby Rudin:

Corollary. In the context of metric spaces) If $F$ is closed and $K$ is compact then $F \cap K$ is compact.

Proof. Because intersections of closed sets are closed and because compact subsets of metric spaces are closed, so is $F \cap K$; since $F \cap K \subset K$, theorem 2.35 shows $F \cap K$ is compact.

He assumes that $F \cap K$ is a compact subset in order to prove $F \cap K$ is compact.

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No, I think he's arguing that $K$ is closed because it is compact; hence $F \cap K$ is closed. –  Asal Beag Dubh Dec 10 '13 at 17:03
ohhhh Thank you. –  Don Larynx Dec 10 '13 at 17:03

That's not what Rudin says. He says that since $\;F\cap K\;$ is closed [as an intersection of closed sets] and $\;F\cap K\subset K\;$ and $\;K\;$ is compact, then so is $\;F\cap K\;$ .

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Just to be clear, at the point where Rudin says "so is $F\cap K$" (right before the semicolon), he is concluding that $F\cap K$ is closed; that's what allows him to invoke Theorem $2.35$, which says that closed subsets of compact sets are compact. –  Barry Cipra Dec 10 '13 at 17:16
Rudin does not say "...so is $\;F\cap K\;$..." at all. I was paraphrasing the book, not quoting it exactly. –  DonAntonio Dec 10 '13 at 17:24
I was quoting the OP, which I assumed was quoting from Rudin. –  Barry Cipra Dec 10 '13 at 17:30

Claim: Compact subsets of metric spaces are closed.

Proof. Suppose the compact set $K$ is not closed. Then there is a limit point $x \notin K$ such that $x_n \in K$ and $x_n \to x$ with respect to the metric.

Consider the open cover which are rings around $x$. That is, consider a sequence $\{r_n\}_{n=-\infty}^\infty$ with $r_n < r_{n+1}$. Let as well $r_n \to 0$ as $n \to -\infty$ and $r_n \to \infty$ as $n \to \infty$. Take $O_n = B_{r_{n+1}}(x)\setminus B_{r_{n-1}}(x)$. Let the open cover be then $K \subseteq \cup_{n=-\infty}^\infty O_n$.

Since $K$ is compact, there is a finite subcover. But then we exclude $O_{m}(x)$ for some smallest $m$, and hence $B_{r_{m-1}}(x) \cap K = \emptyset$. But this contradicts that $x$ is a limit point.

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A sequence from minus infinity to plus infinity? What for?? –  DonAntonio Dec 10 '13 at 17:25
@DonAntonio The rings have to be arbitrarily small and arbitrarily large for it to be an open cover (of the whole space, actually). It's more efficient to index the open sets with respect to the integrs. I think it is pretty clear. –  nayrb Dec 10 '13 at 18:04
I really am not sure what "rings" you're talking about, @nayrb...perhaps you mean open balls? Anyway, I'm not sure it is "pretty clear" that it is mlore efficient (how come?!) to index open sets, or whatever, with the integers and not merely with the naturals. After all they both have the same cardinality. –  DonAntonio Dec 11 '13 at 4:55
$O_n=B_{r_{n+1}}(x) \setminus B_{r_{n-1}}(x)$ are the rings I am talking about. Think donuts in the case of $R^2$ or shells in the case of $R^3$. –  nayrb Dec 11 '13 at 4:57
Ok, I see @nayrb. Thanks. –  DonAntonio Dec 11 '13 at 4:58

I have a Second Edition (1964) of Rudin in which the proof is given this way:

Theorems $2.26(b)$ and $2.34$ show that $F\cap K$ is closed; since $F\cap K \subset K$, Theorem $2.35$ shows that $F\cap K$ is compact.

Theorem $2.26(b)$ says that intersections of closed sets are closed, $2.34$ says that compact subsets of metric spaces are closed, and $2.35$ that closed subsets of compact sets are compact.

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That's exactly what I have. Compare to OP. –  Don Larynx Dec 10 '13 at 18:16
@DonLarynx, it's not how you presented it in the OP. I assumed you were quoting verbatim. –  Barry Cipra Dec 10 '13 at 19:05