I want to prove that

Let $K$ be a compact metric space, Let $S$ be a subset of $C(K)$, prove that if $S$ is uniformly closed, pointwise bounded, and equicontinuous, then $S$ is compact.($C(K)$ is the set of all complex valued, continuous, bounded functions with domain $K$, whose norm is defined as $\|f\|=\sup_{x\in K}|f(x)|$)

the derivation of baby Rudin is following:

By Ascoli's theorem, every $\{g_n\}\subset S$ have a subsequence that converges uniformly, hence converges in the metric of $C(K)$, and limit belongs to $S$, and because of Exercise 26 of Chapter 2 give $S$ is compact.

Ascoli's theorem: If $K$ is compact, if $g_n\in C(K)$, and if $\{f_n\}$ is pointwise bounded and equicontinuous on $K$, then $\{f_n\}$ contains a uniformly convergent subsequence.

and Exercise 26 of Chapter 2: If $S$ be a metric space in which every infinite subset has a limit point, then $S$ is compact.

Question: Ascoli's theorem and closed assumption of $S$ said existence a subsequence which converges on $S$, however exercise 26 of chapter 2 needs the assumption that every infinite subset has a limit point, So I don't know why this proof gives our main question. Is there anyone who can give me some idea to explain this?

  • $\begingroup$ What is a c'pt metric space? $\endgroup$ – copper.hat Sep 15 '17 at 4:06
  • $\begingroup$ compact metric space $\endgroup$ – 백주상 Sep 15 '17 at 4:07
  • $\begingroup$ Yes, every infinite subset having a limit point is equivalent to every sequence having a convergent subsequence. $\endgroup$ – астон вілла олоф мэллбэрг Sep 15 '17 at 4:07
  • $\begingroup$ We may rephrase the condition as being that every countably infinite subset has a limit point, since if a set has cardinality larger than $\mathbb{N},$ we may always take a countable subset, and if this has a limit point, then so did the original subset. Then it is enough to say that every sequence $\{g_{n}\}$ has a convergent subsequence, since this means in particular that every countably infinite set in $S$ has a limit point, as desired. $\endgroup$ – RideTheWavelet Sep 15 '17 at 4:08
  • $\begingroup$ Actually, I didn't learn the notion of cardinality. $\endgroup$ – 백주상 Sep 15 '17 at 4:17

Let $S$ be uniformly closed, pointwise bounded and equicontinuous.

We want to see it is compact, so let $D$ be any countable subset of $S$. Write $D$ as $\{f_n : n \in \mathbb{N}\}$, and we have a sequence $(f_n)_n$ in $D$, so in $S$. Ascoli applies: all subsets of $S$ are equicontinuous and pointwise bounded, so there is a subsequence $(f_{n_k})_k$ that converges to some $f \in C(K)$, uniformly. As all $f_{n_k} \in S$ and $S$ is uniformly closed, we have that $f \in S$, and clearly $f$ is a limit point of $D$ (a sequence from $D$ converges to it). This holds for any countable $D \subseteq S$, so the exercise implies that $S$ is compact.

  • $\begingroup$ What is $(f_n)_n$ and $f_{n_k})_k$? $\endgroup$ – 백주상 Sep 15 '17 at 4:43
  • $\begingroup$ The sequence of $f_n$ and its subsequence. $\endgroup$ – Henno Brandsma Sep 15 '17 at 4:44
  • $\begingroup$ You mean that $f$ is limit point of $D$,but $(f_n)_n$ is not converge to $f$? $\endgroup$ – 백주상 Sep 15 '17 at 4:50
  • $\begingroup$ Ascoli gives us that $f_{n_k} \to f$ ($k \to \infty$). This makes $f$ a limit point of $D$. $(f_n)_n$ need not converge. $\endgroup$ – Henno Brandsma Sep 15 '17 at 4:52
  • $\begingroup$ Thank you, maybe I understand what you mean. $\endgroup$ – 백주상 Sep 15 '17 at 4:52

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