Compact normed vector space Let $V$ be a normed vector space.If $V\neq \{0\}$ is it true that our space cannot be compact?
 A: Here is a compact normed vector space. Let $\mathbb F_2$ be the field with two elements $0,1$. Then $V:=\mathbb F_2$ is a $\mathbb F_2$ vector space. With the norm
$$
\|v\|:= \begin{cases} 0  & v=0\\ 1 & v=1\end{cases}
$$
is becomes a normed vector space. Of course, it is compact.
Seriously, normed vector spaces are usually defined over the reals or complex numbers. Then if $V\ne \{0\}$ there is a non-zero vector $v\in V$. Now consider the sequence $v_n:=n\cdot v$. This sequence does not have a convergent subsequence, hence $V$ cannot be compact.
A: A norm always implies a vector space over the field $\mathscr{F}$ of real or complex numbers. For any unit vector $x \ne 0$,  the set $\{ \alpha x : \alpha \in \mathscr{F} \}$ is isometrically isomorphic to $\mathscr{F}$, which is definitely not compact.
A: A normed real vector space is also a metric space. A metric space is compact if and only if it is complete and totally bounded (in particular, complete and bounded). No non-zero real vector space is bounded  so no such space is compact. A similar argument holds for complex vector spaces.
