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Oct 29, 2020 at 11:39 comment added bof And (assuming $A$ is infinite) the bound is attained by the Cantor space $2^{2^A}$ which has a dense subset with the same cardinality as $A$, e,g., the set of all functions $f:2^A\to2$ such that $f(x)$ depends on only finitely many coordinates of $x\in2^A$.
Oct 29, 2020 at 9:42 history answered YCor CC BY-SA 4.0