Totally bounded, sequentially compact, complete, bounded, closed, equicontinuous $\Rightarrow$ compact?

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)

• 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. – GEdgar Dec 27 '12 at 13:17
• @GEdgar Isn't it common to denote finite as a set equipotent with a finite ordinal in ZF? – Katlus Dec 27 '12 at 13:41
• Just make it clear you don't mean Dedekind finite. Every open cover has a Dedekind finite subcover ... I wonder what those spaces are. – GEdgar Dec 27 '12 at 13:48
• @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! :-) – Asaf Karagila Dec 29 '12 at 9:25

• 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. – Katlus Dec 27 '12 at 13:38
• @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. – Katlus Dec 30 '12 at 17:19
• Camilo's argument seems fine just using upper-bound-property of $\mathbb{R}$math.stackexchange.com/questions/259319/… – Katlus Dec 30 '12 at 17:20