Dense Subspace: Separability

Question

Given a topological space $\Omega$.

Consider a dense subset: $$S\subseteq\Omega:\quad\overline{S}=\Omega$$

Then does it hold true: $$\Omega\text{ separable}\iff S\text{ separable}$$

Clearly it holds: $$\exists A_0\subseteq S\subseteq\Omega:\quad\overline{A_0}=\overline{S}=\Omega\quad(\#A_0\leq\#\mathbb{N})$$ But what about the converse?

The converse is true if we are in the realm of metric spaces (see: http://math.stackexchange.com/questions/1355403/if-x-is-a-separable-metric-space-and-m-subset-x-is-a-metric-subspace-then), but it fails for general topological spaces (see:http://math.stackexchange.com/questions/758424/give-an-example-of-a-non-separable-subspace-of-a-separable-space). — Kolmin, Oct 05 '15 at 20:39
@Kolmin: The second question that you cite isn’t applicable here: the non-separable subspaces in the answers given aren’t dense in the space. — Brian M. Scott, Oct 05 '15 at 20:43

score 6 · Accepted Answer · edited Apr 13 '17 at 12:21

Not necessarily. For each $r\in\Bbb R$ let $D_r=\{0,1\}$ with the discrete topology, and let $X=\prod_{r\in\Bbb R}D_r$. $X$ is separable by the Hewitt-Marczewski-Pondiczery theorem.

For $x\in X$ let $\operatorname{supp}x=\{r\in\Bbb R:x_r=1\}$, the support of $x$. Let $D=\{x\in X:\operatorname{supp}x\text{ is finite}\}$, the set of points of $X$ with finite support. It follows easily from the definition of the product topology that $D$ is dense in $X$. However, $D$ is not separable.

To see this, suppose that $C\subseteq D$ is countable, and let $A=\bigcup_{x\in C}\operatorname{supp}x$; clearly $A$ is countable, so there is an $r\in\Bbb R\setminus A$. Define $x\in X$ by $x(r)=1$ and $x(s)=0$ for $s\in\Bbb R\setminus\{r\}$; clearly $x\in D$, and $\{y\in X:y_r=1\}$ is an open nbhd of $x$ disjoint from $C$, so $C$ is not dense in $D$. Thus, $D$ is not separable.

Note that $X$ is a pretty nice space, since it’s compact and Hausdorff.

Dense Subspace: Separability

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