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.
\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