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Let $X=[0,1]^\omega$. We know that under the standard product topology, $X$ is metrizable. But is it metrizable under the box topology? My guess is not, and the intuition is that Urysohn's metrization theorem fails because the basis for $X$ under the box topology is uncountable. But that's not sufficient to disprove metrizability.

How can one show that this is indeed the case? (or prove otherwise)

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$X$ with the box topology is not even first-countable, which immediately implies it's not metrizable. To see this, suppose we had a countable neighborhood basis $U_1, U_2, \ldots$ at $(0, 0, \ldots)$. Then for each $n$, suppose $[0, \epsilon_{n,1}) \times [0, \epsilon_{n,2}) \times \cdots \subseteq U_n$. Then $V := [0, \frac{1}{2} \epsilon_{1,1}) \times [0, \frac{1}{2} \epsilon_{2,2}) \times \cdots$ is an open neighborhood of $(0, 0, \ldots)$ in the box topology. However, for each $n$, $(0, 0, \ldots, \frac{3}{4} \epsilon_{n,n}, 0, \ldots)$ (with the nonzero entry at position $n$) is in $U_n$ but not in $V$; so no $U_n$ is contained in $V$.

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  • $\begingroup$ Thanks, I didn't know metrizability implies first-countable, but reading the proof makes perfect sense. $\endgroup$ – Yoni Apr 20 '17 at 8:20
  • $\begingroup$ Well, just in case, here's the one-line proof of that fact: if $X$ is a metric space and $x \in X$ then the collection of open balls $B_{1/n}(x)$ is a neighborhood basis at $x$. $\endgroup$ – Daniel Schepler Apr 20 '17 at 16:54
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In the Handbook of Set-theoretic topology (chapter on box products) it is shown that any infinite box product of spaces, that have at least two disjoint open sets (to avoid trivialities) is non-metrisable, in particular not first countable.

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