I'm interested in characterizing closed maps in terms of nets.

Since a map is closed iff $\overline{f(V)} \subseteq f(\overline{V})$ for all subsets $V$, I believe one possible such characterization is

$f$ is closed iff for each net $x_\alpha$ such that $f(x_\alpha) \to y$, we have that $x_\alpha$ converges to the set $f^{-1}({y})$.

By convergence to a set I mean that for any neighborhood $V$ of $f^{-1}({y})$, $x_\alpha$ is eventually in $V$.

In case anyone's interested, I came to this question trying to show that

$X$ is compact $\iff$ the map $X \to \{*\}$ is universally closed.

I know a proof, but I thought a proof in terms of nets could be more elucidatory.


It seems the following.

You belief is right and we can prove a

Proposition. Let $X,Y$ be topological spaces and $f:X\to Y$ be a map. The map $f$ is closed iff for each net $\{x_\alpha\}$ such that $\{f(x_\alpha)\}$ converges to a point $y$, we have that the net $\{x_\alpha\}$ converges to a set $f^{-1}({y})$.

Necessity. Suppose that the map $f$ is closed. Let $\{x_\alpha:\alpha\in A\}$ be a net such that $\{f(x_\alpha)\}$ converges to a point $y\in Y$. Assume that the net $\{x_\alpha\}$ does not converge to a set $f^{-1}({y})$. Therefore there exists an open neighborhood $U$ of the set $f^{-1}({y})$ such that for any index $\beta\in A$ there exists an index $\alpha\in A$, $\alpha\ge\beta$ and $x_\alpha\not\in U$. Put $V=\{x_\alpha: x_\alpha\not\in U\}$. Since the map $f$ is closed, $\overline{f(V)} \subset f(\overline{V})$. Since $\overline{V}\subset X\setminus U\subset X\setminus f^{-1}({y})$, $y\not\in f(\overline{V}) $. But, since $y$ is a limit point of the net $\{f(x_\alpha)\}$, $y\in \overline{f(V)}$, a contradiction.

Sufficiency. Assume that the map $f$ is not closed. Then there exists a (necessarily non-empty) set $V\subset X$ such that $\overline{f(V)}\not\subset f(\overline{V})$. Let $y\in \overline{f(V)}\setminus f(\overline{V})$ be an arbitrary point. Let $A$ be a family of all open neighborhoods of the point $y$. For any neighborhood $U\in A$ pick an arbitrary point $y_U\in f(V)\cap U$ and an arbitrary point $x_U\in V\cap f^{-1}(y_U)$. Then a net $\{x_U: U\in A \}$ converges to a set $f^{-1}(y)$. Since $y\not\in f(\overline{V})$, $f^{-1}(y)\cap \overline{V}=\varnothing$. Thus $X\setminus \overline{V}$ is an open neighborhood of the set $f^{-1}(y)$, which has no points from the set $V$, and, therefore, no points from the net $\{x_U\}$, a contradiction.

  • $\begingroup$ What about a characterization of closed maps in terms of filters, as follows: A map $f: X \to Y$ is closed iff the following condition holds: for any filter $\mathcal{F}$ in $f(X)$ with cluster point $y$, the set $f^{-1}(\{y\})$ is a subset of the cluster points of $f^{-1}(\mathcal{F})$? $\endgroup$ – Eric Auld May 29 '15 at 8:33

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