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Let us say that for a Hausdorff topological space to be locally compact means that every point has a compact neighborhood. Why do locally compact have the property that if $x \in U$ and $U$ is open in $X$, then $x$ has a compact neighborhood that is contained in $U$? If this is not true, which I suspect, then at least it should be true for the spectrum of an abelian C* algebra, as hinted by a proof in Takesaki where he uses this fact without proof. Recall the spectrum is the set of (nonzero) multiplicative linear functionals on a C* algebra. Perhaps the level of generality at which locally compact implies the stronger version of locally compact is in between. Perhaps it holds for some subsets of the dual of a Banach space?

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Just in case: Your title might indicate that you think that the weak$^\ast$-topology on all of a dual Banach space is locally compact. This is not true. Recall that a locally compact topological vector space is finite-dimensional. – t.b. Sep 3 '12 at 10:09
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All of the common definitions of local compactness are equivalent in Hausdorff spaces. In particular, let $X$ be a Hausdorff space in which every point has a compact nbhd. Suppose that $x\in U\subseteq X$, where $U$ is open, let $N$ be a compact nbhd of $x$, and let $V=\operatorname{int}N$; by definition $x\in V$. Now $N$ is a compact Hausdorff space, so it’s regular, and there is a $W$ open in $N$ such that $x\in W\subseteq\operatorname{cl}W\subseteq V\cap U\subseteq N$. Let $G$ be an open set in $X$ such that $W=N\cap G$. Then $W=N\cap G\cap V\cap U=G\cap V\cap U$ is open in $X$. Finally, $\operatorname{cl}W\subseteq N$, and $N$ is compact, so $\operatorname{cl}W$ is compact. That is, $W$ is an open nbhd of $x$ whose compact closure is contained in $U$.

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It follows from the regularity of locally compact Hausdorff spaces. Let say that $N$ is an open neighbourhood of $x$ with compact closure by regularity there is a $V$ so that $x\in V\subset \overline V \subset U\cap N$. So $\overline V$ is a closed subspace of the compact space $\overline N$ hence it is compact.

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