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?
Subsets of the Tychonoff cube.
2 Answers
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 $\mathscr{B}$.
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0I like it. Thank you Brian! – 2012-02-02
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
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.
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0@josh: Ach, you’re quite right: I must have been asleep when I wrote that. The problem, of course, is that the metric in question doesn’t generate the product topology. Why don’t you write up the correct result, including the proof of the corrected proposition; you’ll then be able (after a day or so) to accept your own answer. – 2012-01-31