Characterization of open maps in terms of nets Here I asked about characterization of closed maps in terms of nets/sequences. I find this view illuminating, so I wanted to ask about open maps.
A map $f: X \to Y$ is open if for each $x \in X$ and $U$ neighborhood of $x$, there is a neighborhood $V \ni f(x)$  such that $V \subseteq F(U)$.
We should be able to translate this into a statement about nets, I think.
 A: Proposed alternate characterization:

A map $f:X\rightarrow Y$ is open if for any $(x,y)\in f$ and net $(y_\beta)$ in $Y$ with limit $y$, there is a net $(x_\alpha)$ in $X$ with limit $x$, and such that $f\left(\{x_\alpha\}\right)\subseteq \{y_\beta\}$.

First, let's show that this is sufficient to demonstrate that a map is open in the usual sense (taking open sets to open sets).  Assume that it's not: then despite the hypothesis, there's an open set $U\subseteq X$ such that $f(U)$ is not open, i.e., there is some $y \in f(U)$ such that every neighborhood of $y$ contains a point not in $f(U)$.  Let $x \in U$ be such that $f(x)=y$, and let $(y_\beta)$ be a net in $Y$ indexed by the neighborhoods of $y$ (ordered by inclusion), and such that $y_\beta \in \beta \setminus f(U)$ for each neighborhood $\beta$.  That is, $\{y_{\beta}\}$ and $f(U)$ are disjoint.  Clearly $(y_\beta)$ has limit $y$, since it is eventually in $V$ for any neighborhood $V$ of $y$ (in particular, $y_\beta \in \beta \subseteq V$ whenever $\beta \ge V$).  But for any net $(x_\alpha)$ in $X$ such that $f(\{x_\alpha\})\subseteq \{y_\beta\}$, the sets $f(\{x_\alpha\})$ and $f(U)$ are disjoint; hence $x_\alpha \not\in U$ for all $\alpha$, and therefore $(x_\alpha)$ is not in $U$ eventually (or ever) and cannot have limit $x$.  This is a contradiction, so the hypothesis above is sufficient to establish that $f$ is open.
To show that it is necessary, assume that $f$ is an open map.  Let $(x,y)\in f$ and let $(y_\beta)$ be a net in $Y$ with limit $y$.  Then $(y_\beta)$ is eventually in each neighborhood of $y$.  Because $f$ is open, for each neighborhood $U$ of $x$, there is a neighborhood $V$ of $y$ such that $V\subseteq f(U)$; then $(y_\beta)$ is eventually in $f(U)$, and in particular there is some $\beta(U)$ in the index set of $(y_\beta)$ such that $y_{\beta(U)} \in f(U)$, and hence some $x_U\in U$ satisfies $f(x_U)=y_{\beta(U)}$.  Let $(x_\alpha)$ be a net in $X$ indexed by the neighborhoods of $x$ (ordered by inclusion) such that $x_\alpha\in \alpha$ and $f(x_{\alpha})=y_{\beta(\alpha)}$ for each neighborhood $\alpha$ of $x$.  For each neighborhood $U$ of $x$, $(x_\alpha)$ is eventually in $U$, since $x_\alpha \in \alpha \subseteq U$ whenever $\alpha \ge U$.  We conclude that $(x_\alpha)$ has limit $x$, and since $f(\{x_\alpha\}) \subseteq \{ y_\beta \}$ as well, this completes the proof.
Since we've shown both implications, we have demonstrated that the alternate characterization in terms of nets is equivalent to the usual one in terms of open sets and/or neighborhoods. $\;\square$
