TITLE: Show that X is a separable space if X/U and U are separable and U is closed.
QUESTION [2 upvotes]: Let $X$ be a normed linear space, $U \subset X$ a closed linear subspace.
Both $U$ and $X/U$ are separable.
How can I show that $X$ is separable?
My current approach is as follows:
I have countable subsets $\tilde U = \{u_n\} \subset U$ and $\tilde V = \{y_n + U\} \subset X/U$.
Given those I can define the countable subset $W = \{u_i + y_j\} \subset X$ of which I just need to show that it is dense in $X$.
So given a $x \in X$ I'd like to construct a series $(x_n) \subset \{u_i + y_j\}$ that satisfies $\lim_{n\to\infty} x_n = x$.
There is a series $([y_n]) \subset \tilde V$ which converges against $[x]$, but I am not sure how to continue from there to prove that $W$ is dense.

REPLY [3 votes]: With your notations, take $([v'_i])$ a sequence composed of $[v_j]$s which converges to $[x]$.
The fact that $([v'_i])$ converges to $[x]$ implies that for $m>0$, there exists some $i$ such that
$$\inf_{u\in U}\|v'_i-x+u\|=\inf_{u\in D(U)}\|v'_i-x+u\|\leq\frac1m$$
where $D(U)$ is a countable dense subset of $U$.
So, for $m>0$ there is some $i$ and some $u_m\in D(U)$ such that $\|v'_i+u_m-x\|\leq \frac2m$
This means that that every element of $X$ can be approached by the sum of some element of the countable dense subset of $U$ and any element of $[y]$ for such a class in the countable dense subset of $X/U$. So, $X$ is separable.
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