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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.

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    $\begingroup$ 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). $\endgroup$ – Henno Brandsma Sep 18 '19 at 19:55
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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.

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    $\begingroup$ 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 $Cl_Y(Z)=Y\cap Cl_X(Z)=Y\cap X=Y.$ $\endgroup$ – DanielWainfleet Sep 19 '19 at 0:53
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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 $X$.

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