a question about Riesz's theorem Riesz's theorem:
     Let $(V,\|\cdot\|)$ be a normed vector space, and suppose $C$ is a compact subset of $V$, moreover, $C$'s interior is not empty, and then please prove $\dim(V)<\infty$. 
Because I am not sure whether this statement is true or not, I need to prove it. Actually, I found a Riesz's lemma which states that Let X be a normed vector space and Y be a closed linear subspace of X such that Y does not equal to X and $\alpha$$\in R$,$0<\alpha<1$. Then there is $x_{\alpha}\in X$ such that $\|x_{\alpha}\|=1$ and $\|x_{\alpha}-y\|>\alpha$ for all y$\in Y$, are there any relation between them? How to prove Riesz's theorem? Thanks!
 A: Yes the two are related. Notice that we can assume that $C$ is a closed ball in $V$. If it were not, we could simply pick a closed ball lying in the interior of the set. As the ball is closed in a compact set, it must also be compact. Therefore, we need only show that any closed ball in $V$ is not compact if $\dim V $ is not finite. We may also assume that the ball is the unit ball.
Let $B$ denote the unit ball in $V$. Recall that finite dimensional subspaces of a normed vector space are closed. Let $v_1$ be an element of the ball. By the Riesz lemma, there exists a $v_2$ of unit norm so that $$|| v_2 - y|| > 1/2$$ for any $y$ in the span of $\{v_1\}$. Similarly, there exists a $v_3$ of unit norm so that
$$
|| v_3 - y|| > 1/2
$$
for any $y$ in the span of $\{v_1,v_2\}$. Continuing this procedure, we obtain a sequence $\{v_n\}$.
This sequence does not have any convergent subsequences. Why? Well, for any $n$ and $m$, $||v_n - v_m|| > 1/2$, so the difference of terms is always large.
Finally, we recall the fact that in a metric space any sequence in a compact set must have a convergent subsequence. This completes the proof.
