Normed space $V$ separable $\Leftrightarrow$ unit sphere $\{x\in V: |x|=1\}$ separable

Question

This is my very first post on mathstack even though I have visited here a lot. I think this community is great and "teaches" many things that are not discussed in class, so I must thank all you active advisors here. But to the question. This is my homework problem and I would like to have comments on my proof.

Let $V$ be a normed space. Show that $V$ is separable $\Leftrightarrow$ unit sphere $S=\{x\in V: \lVert x\rVert=1\}$ is separable.

Proof:

"$\Rightarrow$ ": Because $V$ is metric space, then $V$ is separable $\Leftrightarrow$ $V$ is 2nd countable. Therefore $S$ inherits 2nd countability and also being metric space is thus separable.

"$\Leftarrow$": Assume that $S$ is separable. Then it contains countable dense subset $A\subset S$ s.t. $\bar{A}=S$. Now define set $$Q=\bigcup\limits_{q\in\mathbb{Q}}qA,$$ where $qA=\{qa:a\in A\}$. $Q$ is countable union of countable sets and thus countable. We then show that $\bar{Q}=V$. Let $x\in V$. We can assume that $x\neq 0$, because $0\in Q$. Now $$\frac{x}{\lVert x\rVert}\in S.$$ Because $\bar{\mathbb{Q}}=\mathbb{R}$, there exists sequence $q_n\in\mathbb{Q}$ s.t. $q_n\rightarrow \lVert x\rVert$. Also, because $\bar{A}=S$, there exists sequence $a_n\in A$ s.t. $a_n\rightarrow \frac{x}{\lVert x\rVert}$. Then $$q_n a_n\in q_n A\subset\bigcup\limits_{q\in\mathbb{Q}}qA=Q \quad \forall n $$ and $$q_n a_n\rightarrow \lVert x\rVert\cdot \frac{x}{\lVert x\rVert}=x. $$ Therefore $x\in\bar{Q}$ and we are done (?).

Answer 1

Your proofs seem fine. But to make this answer seem less like a comment, I'll include some food for thought:

$(1)$ The proof of the forward implication ($\Rightarrow$) can be done without invoking second countability. Just observe that if $\{x_n\}$ is a countable dense subset of $V$, then $\left\{\frac{x_n}{\|x_n\|}\right\}$ is a countable dense subset of $S$.

$(2)$ The proof of the reverse implication ($\Leftarrow$) can be easily extended to the case of complex scalars.

$(3)$ The only place where I can see some stickler possibly docking points is in the second-to-last line of the reverse implication: $$q_n a_n\rightarrow \lVert x\rVert\cdot \frac{x}{\lVert x\rVert}=x.$$ Your professor may (or may not) prefer you show directly that

There is a countable subset $Q$ of $V$ such that for any $x\in V$ and any $\varepsilon>0$ there is some $a\in Q$ such that $$ \|a-x\|<\varepsilon.$$

Of course, you have all of the ingredients of this, just just invoke continuity of the scalar product.

Nevertheless, fine work. I'd give you an $A$.