Why the weak * topology on the dual of a Banach space has the stronger meaning of locally compact 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?
 A: 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 $U$.
A: 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.
