Let $X$ be a locally compact Hausdorff space. Does the diagonal $\Delta X \subset X \times X$ have a (closed) neighborhood $N$, such that the canonical projection maps $N \to X$ are proper?
1 Answer
Not always.
The first uncountable ordinal $\omega_1$, when given the usual order topology, provides a counterexample.
That this space is locally compact is pretty well known. I claim that no (closed) neighborhood of the diagonal in $\omega_1 \times \omega_1$ has proper canonical projection maps.
(Recall that a proper map is one for which the inverse image of every compact set is compact. Also recall that every compact subset of $\omega_1$ is bounded.)
To see that this claim is true, suppose $U$ is a neighborhood of the diagonal in $\omega_1 \times \omega_1$. For each $\alpha \in \omega_1$, there are some $\alpha^U_0$, $\alpha^U_1$, both strictly less than $\alpha$, such that $(\alpha^U_0,\alpha] \times (\alpha^U_1,\alpha] \subseteq U$ (because sets of this form provide a basis for the topology on $\omega_1 \times \omega_1$). The function $\alpha \mapsto \alpha^U_0$ is regressive, so we may apply Fodor's pressing down lemma: there is some $\gamma \in \omega_1$ and a stationary $S \subseteq \omega_1$ such that $\alpha^U_0 = \gamma$ for all $\alpha \in S$. This means that $$\{\delta \in \omega_1 : (\gamma+1,\delta) \in U\}$$ is uncountable (it contains $S$), hence not compact. But this set is also the inverse image of $\{\gamma+1\}$ under the canonical projection map onto the first coordinate. Thus this projection is not perfect.