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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.

Can you give me an insight about this?

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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.

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  • $\begingroup$ 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? $\endgroup$ – user8469759 Nov 7 '18 at 16:00
  • $\begingroup$ Pointing this out because the statement of Baire's theorem seems to be valid for "locally compact Hausdorff spaces". $\endgroup$ – user8469759 Nov 7 '18 at 16:01
  • $\begingroup$ $K$ is a subset of a Hausdorff space (namely $X$), hence Hausdorff, and compact implies locally compact. $\endgroup$ – Aweygan Nov 7 '18 at 16:22
  • $\begingroup$ The question might appear silly, why does compact imply locally compact? $\endgroup$ – user8469759 Nov 7 '18 at 16:28
  • $\begingroup$ 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. $\endgroup$ – Aweygan Nov 7 '18 at 16:29

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