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Having inspected the proof of the fact that for every compact metric space $X$ there exists a continuous surjection from $\{0,1\}^{\mathbb N}$ onto $X$, it looks to me it is a theorem of ZF + DC (dependent choice).

Is there any reference for this fact in the literature?

Tomasz Kania
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    I'd start by looking at the works of Eleftherios Tachtsis, Kyriakos Keremedis, and their coauthors. Or in Eric S. Schechter's "Handbook of Analysis and its Foundation" which is chock-full of choice-related information. Skimming through the outline of the proof (e.g. https://math.stackexchange.com/questions/1873116/), it seems that separability of compact metric spaces might be enough, and this is equivalent (if my memory serves me right) to countable choice (which is weaker than DC). But this is just shooting from the hip. – Asaf Karagila Apr 30 '20 at 11:46
  • @AsafKaragila, thank you Asaf. Yes, the proof deals with a countable base of a compact space and after all it boils down to $\textsf{AC}_\omega$. Thank you for the suggestions! – Tomasz Kania Apr 30 '20 at 11:48

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The proof I would use, is to use that $X$ has a countable base (this needs countable choice, I think), and so $X$ embeds into $[0,1]^\omega$, quite explicitly. And as the Cantor function $c: C \to [0,1]$ maps $C$ onto $[0,1]$ (also choicelessly), $C$ maps onto $[0,1]^\omega$ too, with $c'=c^\omega$ essentially. Then Kechris in his book shows that any closed subset of $C$ is a retract of it (using branches in a finitary tree argument, I recall it as being quite explicit), so $c'^{-1}[X]$ is too, and hence $X$ is an image of $C$ (composing the retract with $c'$). Cursorily I only see the Urysohn embedding (from a countable base) as the part that would need countable choice.

Henno Brandsma
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  • Thank you, Henno (good to see you again!). Yes, I think I can prove it myself but was more interested in a citable source if it exists. – Tomasz Kania May 04 '20 at 19:49