In a textbook I'm reading, I have a topological space $X$ (which is a subspace of $L^{\infty}(E) \setminus \{0\}$ for some topological space $E$), such that $X \cup \{0\}$ is weak*-compact (in this case $L^{\infty}(E)$ is seen as the dual of $L^1(E)$).
Then, they say: "therefore, $X$ is locally compact". Apparently, $X$ is not a closed subspace of $X \cup \{0\}$. I don't understand why they conclude that $X$ is locally compact. I can give more context if necessary.
Thank you for your comments!
An open subset $O$ of a (completely regular, but that's no problem here, all topological vector spaces are) locally compact space $X$ is locally compact.
This follows because for Hausdorff locally compact spaces, being locally compact is equivalent to having a base of open neighbourhoods with compact closure, and any point in $O$ will have that as well (just make sure we only consider those neighbourhoods whose closure lies in $O$, which we can always do by regularity).