Reading this question I learned that a Banach space is $\sigma$-compact iff finite dimensional. The $\impliedby$ direction is obvious, but I cannot find a proof of the $\implies$ direction, I checked Conway, Brezis and DiBenedetto, but they barely mention $\sigma$-compactness.
How do you prove this fact or where can I find a proof of it?
On
Suppose $X$ is a $\sigma$-compact infinite-dimensional Banach space, with $X=\cup_n K_n$. By the Baire category theorem, there exists some $n\in\mathbb N$ such that $K_n$ has nonempty interior. But compact sets in infinite-dimensional normed spaces always have empty interior.
To prove the claim made in this last sentence, assume $X$ is an infinite-dimensional normed space, and $K\subset X$ is compact and has nonempty interior $K^\circ$. Since scaling and translation are homeomorphisms in normed spaces, we may assume without loss of generality that $K$ contains the open ball of radius $>1$ centered at $0\in X$. Then applying induction and the Riesz lemma, we obtain a sequence $(x_n)$ in $K$ with no convergent subsequence, contradicting the fact that $K$ is compact.
On
From Chapter 25 of Willard's General Topology, we have the following nice theorem:
A $\sigma$-compact space $X$ is Baire iff the set of points at which $X$ is locally compact is dense in $X$.
You can combine this with the fact that a topological vector space is locally compact iff it is finite-dimensional to prove that any $\sigma$-compact Baire topological vector space (including any $\sigma$-compact Banach space) must be finite-dimensional (since a topological vector space is locally compact iff it is locally compact at one point).
Hint: First show that in infinite dimension compact sets have no interior points (recall that the unit ball is not compact). Now assume that $X$ is $\sigma$-compact and pick a family $(A_n)_{n\geq 1}$ of compact sets such that $$ \bigcup_{n\geq 1} A_n = X.$$ Now $X$ has an interior point. Use Baire's Theorem to show that this leads to a contradiction.