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Related; When $K$ is compact, if $S\subset C_b(K)$ is closed,bounded and equicontinuous, then $S$ is compact? (ZF)

I just edited my whole question since i think it was a bit messy.

Here is my question.

Let $K$ be a separable compact metric space and $S\subset C(K,\mathbb{C})$.

Let $S$ be closed,bounded,uniformly equicontinuous on $K$, sequentially compact, totally bounded and complete.

Then is $S$ compact? (in ZF)

Thank you in advance!

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    $\begingroup$ Of course, if you do not assume AC, you should say what "compact" means. If it involves the word "finite" you should say what that means. $\endgroup$ – GEdgar Dec 27 '12 at 13:17
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    $\begingroup$ @GEdgar Isn't it common to denote finite as a set equipotent with a finite ordinal in ZF? $\endgroup$ – Katlus Dec 27 '12 at 13:41
  • $\begingroup$ Just make it clear you don't mean Dedekind finite. Every open cover has a Dedekind finite subcover ... I wonder what those spaces are. $\endgroup$ – GEdgar Dec 27 '12 at 13:48
  • $\begingroup$ @GEdgar: It is the common terminology that "finite" means "smaller than $\aleph_0$", so there is no actual confusion. As for your question, assuming the axiom of choice - those are the compact spaces! :-) $\endgroup$ – Asaf Karagila Dec 29 '12 at 9:25
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You might want to read an answer to a slightly different question over at MathOverflow.

The short answer is: Arzelà–Ascoli does require (a weak form of) choice.

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  • $\begingroup$ But if you assume $K$ to be separable, then Arzela Ascoli Theorem does not require choice. So i did assume that $K$ to be separable. $\endgroup$ – Katlus Dec 27 '12 at 13:38
  • $\begingroup$ @Katlus: Are you sure? Baire Category Theorem is true for separable metric spaces, but I'm not sure about Arzela-Ascoli. For more details see Herrlich's book chapter 4.9 (in particular p.99 and onwards). $\endgroup$ – Asaf Karagila Dec 29 '12 at 9:51
  • $\begingroup$ @Asaf Yes, I am sure that "If $K$ is separable compact metric space and $f_n\in C(K,\mathbb{K})$ and $\{f_n\}$ is bounded and equicontinuous on $K$, then $\{f_n\}$ contains a uniformly convergent subsequence" is true under ZF. $\endgroup$ – Katlus Dec 30 '12 at 17:19
  • $\begingroup$ Camilo's argument seems fine just using upper-bound-property of $\mathbb{R}$math.stackexchange.com/questions/259319/… $\endgroup$ – Katlus Dec 30 '12 at 17:20

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