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Let $X$ be a compact metric space. I see why all open and closed subsets of $X$ are separable. But is every subset of $X$ necessarily separable?

EDIT: Since $X$ is separable metric, it embeds into the Hilbert cube $[0,1]^\omega$, which is hereditarily separable, right? And so $X$ is also hereditarily separable.

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  • $\begingroup$ If I recall correctly, every separable metric space is second countable. That should do it. $\endgroup$ – Pedro Sánchez Terraf Feb 11 '16 at 3:18
  • $\begingroup$ Countable Choice turns out to be necessary and sufficient. See here and here. $\endgroup$ – Cameron Buie Feb 11 '16 at 3:37
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Yes: a compact metric space is second countable, and second countability is hereditary and implies separability.

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  • $\begingroup$ Once upon a time, 3 seconds before... $\endgroup$ – Pedro Sánchez Terraf Feb 11 '16 at 3:19

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