We know $\{0,1\}^\mathbb{N}$ is sequentially compact, but $\{0,1\}^\mathbb{R}$ is not. The proof I have seen that the second is not sequentially compact heavily relies on the fact that we have a continuum. So not assuming the continuum hypothesis, is it still true that $\{0,1\}^A$ is not sequentially compact for any uncountable $A$?
2026-03-25 09:34:55.1774431295
Is $[0,1]^A$ not sequentially compact for any uncountable $A$?
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