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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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    @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! :-)2012-12-29

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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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    Camilo's argument seems fine just using upper-bound-property of $\mathbb{R}$http://math.stackexchange.com/questions/259319/is-it-possible-to-choose-a-subsequence-countable-times-in-zf2012-12-30