Proof that no normed space is compact

Question

The notes for my functional analysis class casually state that every nontrivial normed space is not compact, but do not give a proof. I can't tell if this is because it is a trivial proof or because it is too complicated to be given inline. Is there a good proof for this? I have searched around the internet for a while and haven't been able to find one.

Answer 1

Of course, the trivial normed space $\{0\}$ with norm $\|0\|=0$ is compact. So let's restrict our attention to normed spaces with dimension $\geq1$.

Let $X$ be a nontrivial normed space, and let $x\in X$ be nonzero. Now consider the sequence $\{nx:n\in\mathbb N\}$. For each $n$ the set $$B_n=\{y\in X:\|nx-y\|<\frac{1}{2\|x\|}\} $$ is an open neighborhood of $nx$ which does not contain any other point in the sequence. Thus no subsequence is convergent (why?). Hence $X$ is not sequentially compact, and therefore $X$ is not compact.

Answer 2

Any compact subset of any metric space is bounded.

Since any nonzero normed space is unbounded, it isn't compact.

EDIT

Consider any non zero Vector $x$ and any $M>0$; we have $\Vert tx\Vert=\vert t\vert\Vert x\Vert$ which Will be $>M$ for suitable $t$

Answer 3

Let $(X, \|•\|) $ be a normed linear space.

Claim : $X$ is compact iff $X=\{0\}$

Proof: One direction is fairly trivial.

Let us assume that $X$ is compact. Then $X$ must be bounded i.e $\exists r>0$ such that $\|x\|\le r,\forall x\in X$.

For given any $\epsilon>0$ , $\|\epsilon x\|\le r$ as $\epsilon x\in X$ .

Then $0\le \|x\|=\|\frac{\epsilon x}{\epsilon}\|\le \frac{r}{\epsilon}$

Since $\epsilon>0$ is arbitrary, $\|x\|=0$ and hence $x=0$