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?



Does any locally compact space have a proper diagonal neighborhood?
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](https://en.wikipedia.org/wiki/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](https://en.wikipedia.org/wiki/Fodor%27s_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.


