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

Let $S$ be closed,bounded and equicontinuous. The usual proof for this is, using Arzela-Ascoli Theorem and Axiom of countable choice, showing that $S$ is limit point compact. (The problem is, under ZF, additional hypothesis "$K$ is separable" should be added to make Arzela-Ascoli Theorem true)

I think, in order to prove $S$ is compact without AC, one should start directly with an infinite subcover.

Assuming above hypotheses, I have proved $S$ is totally bounded and complete, and it really does seem provable that $S$ is compact under ZF.

Is it unprovable? Or if there is an argument proving this without choice please let me know help!

Thank you in advance

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    I didn't intend it, but since i edited my recent post, now this is exactly a duplicate of my recent post. Please close this post if one sees this. Thank you..2012-12-27
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    http://math.stackexchange.com/questions/265802/totally-bouded-sequentially-compact-complete-bounded-closed-equicontinuous-righ2012-12-27
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    Voting to close, as requested.2013-06-08

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