I wonder if there is a compact and locally Hausdorff space $X$ which is not locally compact, in the sense that every point has a neighborhood base consisting of compact sets.
A space is called locally Hausdorff if every point has a neighborhood which is a Hausdorff space under the subspace topology. Such a space need not be Hausdorff. However, if each point has a closed Hausdorff neighborhood, then the space is Hausdorff. An example is the real line with two origins. If one restricts this space between $-1$ and $1$, it becomes compact. Unfortunately, it is also locally compact.
If $X$ is Hausdorff and compact, then it is also locally compact, so the Hausdorff neighborhoods must all be proper subsets of $X$.
Does anyone have an idea?
Best Answer
Here’s a counterexample to the conjecture.
Let $Y=\Bbb N\times\Bbb N$, let $p$ and $q$ be distinct points not in $Y$, and let $X=Y\cup\{p,q\}$. Points in $Y$ are isolated. For each $k\in\Bbb N$ the set
$$B_p(k)=\{p\}\cup\left\{\langle m,n\rangle\in Y:n\ge k\right\}$$
is a basic open nbhd of $p$. For each $k\in\Bbb N$ the set
$$B_q(k)=\{q\}\cup\left\{\langle m,n\rangle\in Y:m\ge k\right\}$$
is a basic open nbhd of $q$. For convenience, for $n\in\Bbb N$ let $S_n=\{n\}\times\Bbb N$.
Each open nbhd of $p$ contains all but finitely many points of each $S_n$, and each open nbhd of $q$ contains all but finitely many of the sets $S_n$. Thus, every open set containing both $p$ and $q$ contains all but finitely many points of $Y$, and $X$ is compact. $Y$ is Hausdorff, as are the sets $B_p(k)$ and $B_q(k)$, so $X$ is locally Hausdorff. However, none of the sets $B_p(k)$ contains a compact nbhd of $p$, so $X$ is not locally compact. (Similarly, $X$ fails to be locally compact at $q$.)