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Define $H=\{(x_n)_n\in\ell^2:|x_n|\le \frac1n, n\in\mathbf N\}\subset\ell^2$. This set is known as the Hilbert cube and it is well-known that $H$ is compact, convex and non-empty. Let $\overline{\mathrm{conv}}(C)$ denote the closure of the convex hull of a subset $C\subset\ell^2$.

Suppose $S$ is a non-empty, compact, convex subset of $\ell^2$, is it possible to write$$S=\overline{\mathrm{conv}}\left(\bigcup_{n=1}^\infty[ S\cap(n\cdot H)]\right),$$ where (for $n\in\mathbf N$ fixed) $n\cdot H=\{n\cdot x:x\in H\}$.

I think it is possible (since the Hilbert cube keeps getting 'thinner' in each coordinate), but I do not know how to prove it.

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up vote 4 down vote accepted

Take $S:=\{x\}$, where $x_k=\frac{\ln k}k$. Then $S\cap nH$ is empty for all $n$, since if $S\cap nH\subset S$, the only candidate is $x$, and we can't have $\ln k\leq n$ for all $k$.

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More generally: by the Baire Category Theorem, a complete metric space is not the union of countably many closed nowhere-dense sets; a compact set in an infinite-dimensional Banach space is nowhere dense. – Robert Israel Aug 6 '12 at 18:03

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