# Theorem 2.9 rudin functional analysis (application of Baire's theorem)

I'm reading through the theorem 2.9 (Rudin functional analysis) which states

Suppose $$X$$ and $$Y$$ are topological vector spaces, $$K$$ is a compact convex set in $$X$$, $$\Gamma$$ is a collection of continuous linear mapping of $$X$$ into $$Y$$, and the orbits $$\Gamma(x) = \left\{ \Lambda x : \Lambda \in \Gamma \right\}$$ are bounded subsets of $$Y$$, for every $$x$$ in $$K$$. Then there's a bounded set $$B \subset Y$$ such that $$\Lambda(K) \subset B$$ for every $$\Lambda \in \Gamma$$

The proof goes as follows:

Let $$B$$ be the union of all sets $$\Gamma(x)$$, for $$x \in K$$. Pick balanced neighborhoods $$W$$ and $$U$$ of $$0$$ in $$Y$$ such that $$\bar{U} + \bar{U} \subset W$$ put

$$E = \bigcap_{\Lambda \in \Gamma} \Lambda^{-1}(\bar{U}).$$ If $$x \in K$$, then $$\Gamma(x) \subset nU$$ for some $$n$$, so that $$x \in nE$$. Consequently, $$K = \bigcup_{n=1}^{\infty}(K \cap nE).$$ Since $$E$$ is closed, Baire's theorem shows that $$K \cap nE$$ has non empty interior (relative to $$K$$) for at least one $$n$$.

I don't understand how the Baire's theorem is actually applied here, the hypothesis to apply such theorem we need to have a set $$S$$ that is either a complete metric space (which I don't see anywhere) or a locally compact Hausdorff space. I guess somehow the latter is exploited, but I don't see how.

Recall that, by Rudin's definition, all topological vector spaces are Hausdorff. The Baire category theorem is applied to $$K$$, which is by definition compact, hence compact Hausdorff.
• Isn't "locally compact" different from "compact"? I.e. don't I need to show we have a local base whose members have compact closure? Here $K$ is a compact convex set, isn't necessary a topological vector space, right? – user8469759 Nov 7 '18 at 16:00
• $K$ is a subset of a Hausdorff space (namely $X$), hence Hausdorff, and compact implies locally compact. – Aweygan Nov 7 '18 at 16:22
• A space $Z$ is locally compact if each point in $Z$ has a neighborhood with compact closure, and if $Z$ is compact, then it is a neighborhood of each of its points with compact closure. – Aweygan Nov 7 '18 at 16:29