Subsets of the Tychonoff cube. Let $X$ be the Tychonoff cube $X = I^{2^{\omega}}$, where $I = [0,1]$. In class, we saw that $X$ is separable. Would $X$ contain a dense metrizable subset as well?
 A: Added: Of course what follows is nonsense, because the metric does not generate the product topology. As the OP points out in the comments below, a dense subspace of $X$ cannot be first countable.
It even contains a countable dense metrizable subset.
Take $J=[0,1)$ as the index set of the product, so that as a set $X$ is the set of functions from $J$ to $I$. Let $Q=\mathbb{Q}\cap I$. For each $n\in\mathbb{Z}^+$, each $(n+1)$-tuple $\bar q=\langle q_1,\dots,q_{n+1}\rangle\in Q^n$ such that $q_1<\dots<q_{n+1}=1$, and each $n$-tuple $\bar r=\langle r_1,\dots,r_n\rangle\in Q^n$ let $x(\bar q,\bar r)\in X$ be defined by $$x(\bar q,\bar r)_j=r_k\text{ iff }q_k\le j<q_{k+1}\;,$$ and let $D$ be the set of such $x(\bar q,\bar r)$. $D$ is a countable dense subset of $X$. (This or something very similar is probably the most common construction of such a set.)
For any $x,y\in D$, the set $\{|x_j-y_j|:j\in J\}$ is finite, so it has a largest member; denote that member by $d(x,y)$. It’s not hard to check that $d$ is actually a metric on $D$. To check the triangle inequality, for example, suppose that $x,y,z\in D$. Let $j\in J$ be such that $d(x,y)=|x_j-y_j|$. Then $|x_j-y_j|\le|x_j-z_j|+|z_j-y_j|\le d(x,z)+d(z,y)$. Everything else is completely straightforward.
By the way, since $D$ has no isolated points, it must be homeomorphic to $\mathbb{Q}$ with the usual topology, since up to homeomorphism that is the only countably infinite metric space without isolated points.

Added: Here’s a brief sketch of the correct argument: if $D$ is a dense subspace of $X$, $x\in D$, and $\{B_n:n\in\omega\}$ is a countable local base at $x$ in $D$, then $\{\operatorname{int}_X\operatorname{cl}_X B_n:n\in\omega\}$ is a countable local base at $x$ in $X$, since $X$ is regular. But $X$ is nowhere first countable, so $D$ must be nowhere first countable and hence cannot be metrizable. I’ll leave it to the OP to write up and accept a more detailed version if he wishes, since he really answered his own question.
A: Hopefully my proof is correct.
The Tychonoff cube $X$ will not have a dense metrizable subset. To see this, we need the following proposition.
Proposition  Let $X$ be regular and $A$ be a dense subset of $X$. If $A$ is first-countable at a point $x \in A$, then $X$ is first-countable at that same point $x$.
Proof  Let $A$ be a dense subset of a regular space $X$ and let A be first countable at the point $x$. Consider the countable local base $\{ B_n \}$ of $x$ in $A$.
Since $A \subset X$, we have $B_n = V_n \cap A$, where $V_n$ is an open subset of $X$. Therefore, $x \in \overline{B_n} = \overline{V_n \cap A} = \overline{V_n}$. (All closures are taken in $X$.)
We now show $\operatorname{Int}(\overline{B_n}) = \operatorname{Int}(\overline{V_n}) $ will form a base of $x$ in $X$.
By regularity, for any open neighborbood $O_x$ of $x$ (in $X$), we can find an open subset $O^*_x$ such that $x \subset O^{*}_x \subset \overline{O^{*}_x} \subset O_x$. Fix $n$ such that $B_n\subset O^*_x\cap A$; then $\overline{V_n}=\overline{B_n} \subset \overline{O^{*}_x} \subset O_x$, which implies that $\operatorname{Int}(\overline{B_n}) \subset O_x$.
Thus, sets of the form $\operatorname{Int}(\overline{B_n}) = \operatorname{Int}(\overline{V_n}) $ will form a base of $x$ in $X$ and the proposition is proved. $\square$
However, the Tychonoff cube does not have a countable base at any point, so $A$ could never be first-countable and hence cannot be metrizable.
Also note that for a regular space $X$, a dense subset $A$ and a base $\mathscr{B}$ for $A$,
the sets of the form $\operatorname{Int}(\overline{U})$, where $U \in \mathscr{B}$, need not form a base for $X$. For instance, if $X = \mathbb{R}$, $A = \mathbb{Q}$, and $\mathscr{B}$ is the set of all intervals in $\mathbb{Q}$ whose closure (in $\mathbb{R}$) does not contain $\sqrt{2}$, you get a counterexample. The point is that $U$ doesn't have to belong to the given base $\mathscr{B}$.
