Is my solution flawed? (Showing compactness)

Question

Let $(V,||.||)$ be a normed vector space and suppose $S:=\{s\in V:||s||=1\}$ is compact then is $V$ compact as well? Here we mean as sequentially compact.

My attempt:

Define the function $f:V\to S$ as follows:

$ f(x)= \begin{cases} \frac{x}{||x||}&\text{if}\, x\neq 0\\ some s\in S&\text{otherwise} \end{cases} $

I should clarify that I meant to map the zero vector to some vector in $S$ (however I am not entirely comfortable in doing so)

Note that $f(V)=S$ and $f$ is continuous on $V$. Then suppose $V$ is not compact then $S$ is not compact, contradicting our assumption and hence $V$ is compact?

My concern:

My problem lies in mapping the zero vector into some other vector, if I map it into the zero vector then first it is not continuous and second $f(0)\notin S.$ How can I overcome this problem? Can I just map the zero vector into some random vector in $S$ whilst still ensuring its well-definedness?

Answer 1

There are two problems there:

1. The map $f$ is not continuous at $0$, no matter what $s$ is.
2. Just because $f(V)$ is compact, that doesn't prove $V$ is compact. Otherwise, every topological space would be compact; you would just take a constant map $f$.