# Showing that a dense subspace $Y$ of a first countable separable topological space is separable

Show that a dense subspace $$Y$$ of a first countable separable topological space $$X$$ is separable.

Proof:

$$X$$ is separable. Let $$S=\{x_n \in X | n \in \mathbb{N}\}$$ be a countable dense subset of $$X$$.

$$Y$$ is also dense in $$X$$.

Because $$X$$ is first-countable, thus for each $$x_n$$ where $$n \in \mathbb{N}$$ there exists a countable local-basis around $$x_n$$. Let the countable local-basis around $$x_n$$ be $$S_n=\{\text{ }B_n^k \text{ } | \text{ }k \in \mathbb{N} \}$$

Because $$Y$$ is dense in $$X$$ thus for each $$x_n$$ where $$n=1,2,3 \dots$$ and for each $$B_n^k$$ where $$k=1,2,3,4 \dots$$, we have $$Y \cap B_n^k \neq \phi$$.

Say $$y_n^k \in Y \cap B_n^k \neq \phi$$

Denote $$Z=\{ y_n^k \in Y \text{ } | \text{ } n,k \in \mathbb{N} \}$$

Claim: $$Z$$ is a countable dense set of $$Y$$.

Choose $$y \in Y$$ and any open set $$V$$ in $$Y$$ containing y. $$V$$ is open in $$Y$$ implies that $$V=U \cap Y$$ where $$U$$ is an open set in $$X$$.

Thus $$y \in U \in \tau$$ and $$y \in Y$$

$$y \in U$$ and $$U$$ is open in X. Because $$S$$ is dense in X, we have that $$U \cap S \neq \phi$$.

Let $$x_n \in U \cap S$$, Thus $$x_n \in U$$ and $$U$$ is open in $$X$$.

Considering that $$S_n$$ is a countable local-basis around $$x_n$$ we have an element $$B_n^{k_0}$$ such that $$x_n \in B_n^{k_0} \subset U$$. choose the corresponding $$y_n^{k_0}$$ as done in the construction above. Then we have $$y_n^{k_0} \in B_n^{k_0} \subset U$$. Thus $$y_n^{k_0} \in U \cap Y = V$$ and hence $$V \cap Z \neq \phi$$ as it contains $$y_n^{k_0}$$.

Hence $$Y$$ has a countable dense subset. $$Y$$ is separable.

Hence proved!

Please check my solution. I need to correct my mistakes and learn. Thank You.

• Minor MathJax point: \emptyset is more common, \phi is a more "Slavic" tradition notation for the empty set (I've also seen \Lambda used for it). Sep 18, 2019 at 19:55

This proof looks fine. Quite detailed. See Daniel's comment for an alternative faster proof.

To see that you need the first countable assumption on $$X$$: if $$X=[0,1]^\mathbb{R}$$, then $$X$$ is separable (but not first countable), and $$Y=\Sigma_0[0,1]^\mathbb{R} := |\{f \in X: |\{x: f(x) \neq 0\}| \le \aleph_0 \}$$ is dense in $$X$$ and not separable. Think about it.

• For each $n$ let $Y_n=\{y_n^k: k\in \Bbb N\}.$ Then $x_n\in Cl_X(Y_n)$ because $Y_n$ intersects every member of $S_n$ and $S_n$ is a local base at $x_n$..... So $Cl_X(Z)\supset Cl_X(S)=X....$ So $Cl_Y(Z)=Y\cap Cl_X(Z)=Y\cap X=Y.$ Sep 19, 2019 at 0:53

Looks good to me. Here's a slightly different way to look at it. Consider the following properties of a topological space $$X$$:

(1) $$X$$ is separable and first countable;

(2) $$X$$ has a countable $$\pi$$-base, i.e., a countable collection $$\mathcal B$$ of nonempty open sets such that every nonempty open set contains a member of $$\mathcal B$$ as a subset;

(3) $$X$$ is separable.

You showed that a dense subspace of a space with property (1) has property (3). With the same ideas you can show that $$(1)\implies(2)\implies(3)$$, and that a dense subspace of a space with property (2) has property (2).

$$(1)\implies(2)$$: Suppose $$X$$ is separable and first countable. Let $$S$$ be a countable dense subset of $$X$$, for for each $$x\in S$$ let $$\mathcal B_x$$ be a countable local base at $$x$$. Then $$\bigcup_{x\in S}\mathcal B_x$$ is a countable $$\pi$$-base for $$X$$.

$$(2)\implies(3)$$: Suppose $$\mathcal B$$ is a countable $$\pi$$-base for $$X$$. By choosing one point from each member of $$\mathcal B$$, we get a countable dense subset of $$X$$.

Finally, suppose $$X$$ has property (2) and $$Y$$ is a dense supspace of $$X$$. Let $$\mathcal B$$ be a countable $$\pi$$-base for $$X$$; then $$\mathcal B_Y=\{B\cap Y:B\in\mathcal B\}$$ is a countable $$\pi$$-base for $$X$$.