Given $\epsilon > 0$ and $f : [0, 1]^{\omega} \rightarrow [0, 1]^{\omega}$, can we find $x$ such that $x \in \textrm{Conv}\left( \left\{f(y) : ||y - x||_{\infty} < \epsilon\right\}\right)$?

In finite dimensions this is straightforwardly equivalent to Brouwer's fixed point theorem. In the infinite dimensional case they are not equivalent, and I don't even know whether to expect the claim to be true or false (I was expecting it to be much more straightforward to settle).

I would actually be happy to solve the problem for the simpler case of maps $f : [0, 1]^{\omega} \rightarrow \left\{0, 1\right\}^{\omega}$, since I'm imagining each coordinate of $f$ as returning the truth of a certain predicate applied to $x$. And we could just as well work with the discrete setting, of $f : \left\{0, 1, \ldots, m\right\}^{\omega} \rightarrow \left\{0, m\right\}^{\omega}$.

I'm aware of this result, but it doesn't seem to help very much here.

Edit: I would actually be happy to find an $x$ which isn't separated from $\left\{ f(y) : ||y - x||_{\infty} < \epsilon\right\}$ by any hyperplane parallel to the coordinate axes. This seems like it should be much easier.

bethe image of one of its neighbors, so we definitely need to take the convex hull. $\endgroup$ – Paul Christiano Oct 25 '13 at 1:34