Let $X$ be a locally compact Hausdorff topological space.
By locally compact we mean that every $x\in X$ admits a local basis consisting of compact neighbourhoods.
Then since compact Hausdorff spaces are normal, each $x\in X$ admits a local basis consisting of normal neighbourhoods.
My question is whether it must be the case that each $x\in X$ admits a local basis consisting of normal open neighbourhoods.
An issue preventing this from being a completely straightforward deduction is the existence of compact Hausdorff spaces which are not completely normal, e.g. the Tychonoff plank.
Best Answer
Yes: let $x \in U$, $U$ open in $X$. Then there is a compact neighbourhood $W \subset U$ of $x$. Since $X$ is locally compact (and Hausdorff), it is completely regular. Hence there is a cozero set $V$ such that $x \in V \subset W$. Of course, $V$ is open. As a cozero set, $V$ is $F_\sigma$ in $X$ and also in $W$. Since $W$ is normal, and $F_\sigma$ subsets of normal spaces are normal (see, for instance, Engelking, General topology, 2.1.E), $V$ is normal.