# Proper maps and convergent subnets

Let $X$ and $Y$ be Hausdorff topological spaces and let $p: X \rightarrow Y$ be a proper (every compact set has compact preimage) continuous map.

1. Is the following true? Given a net $(x_i)$ in $X$ such that $(p(x_i))$ converges, then $(x_i)$ has a convergent subnet.

2. If so, is there a standard reference?

No, this is not true in general. For instance, let $X$ be an infinite discrete space and let $F$ be a nonprincipal ultrafilter on $X$ (or more generally, a filter that contains the cofinite filter and is not contained in the cofinite filter on any infinite subset of $X$). Let $Y=X\cup\{\infty\}$, where $X$ is an open subspace of $Y$ and neighborhoods of $\infty$ are sets of the form $\{\infty\}\cup A$ for $A\in F$. Note that no infinite subset of $Y$ is compact, since if $S\cup\{\infty\}$ is compact then every element of $F$ must be cofinite in $S$.
It follows that the inclusion map $p:X\to Y$ is proper. But we can choose a net $(x_i)$ in $X$ which converges to $\infty$ in $Y$, and this net has no convergent subnet in $X$.
However, it is true if $Y$ is locally compact. I don't know of a reference off the top of my head, but here's a proof. Let $y$ be the limit of $(p(x_i))$ and choose a compact neighborhood $K$ of $y$. Then $p(x_i)$ is eventually in $K$, so $x_i$ is eventually in $p^{-1}(K)$. Passing to a subnet, we may assume $x_i\in p^{-1}(K)$ for all $i$. But $p^{-1}(K)$ is compact since $p$ is proper, so $(x_i)$ must have a convergent subnet.
Actually, it is true more generally as long as $p$ is a closed map (this is automatic if $Y$ is locally compact). Indeed, suppose $(p(x_i))$ converges to a point $y$. Note that an accumulation point of $(x_i)$ is just an element of the intersection $\bigcap_j \overline{A_j}$ where $A_j=\{x_i:i\geq j\}$ for each $j$ in the index set of the net. If $p$ is closed, then $p(\overline{A_j})$ is closed for each $j$, and thus must contain $y$ since it contains $p(x_i)$ for all $i\geq j$. Thus $B_j=\overline{A_j}\cap p^{-1}(\{y\})$ is nonempty for each $j$. Each $B_j$ is closed in $p^{-1}(\{y\})$, and they have the finite intersection property since the index set is directed. Since $p^{-1}(\{y\})$ is compact, $\bigcap B_j$ is nonempty. Thus $\bigcap\overline{A_j}$ is nonempty, so $(x_i)$ has an accumulation point and hence a convergent subnet.
• Actually, it is true as long as $p$ is closed (which you really usually want to include in the definition of "proper" for more general spaces); I've updated my answer. – Eric Wofsey Aug 10 '16 at 6:38