Nets

For my own personal use, we review the basic theory of nets.

Definition. Fix a topological space \(X\). A net is a sequence \(\{x_i\}_{i\in I}\) indexed by a nonempty directed set \(I\). The net is said to converge to a point \(x\in X\) if and only if each open neighborhood \(U\) of \(x\) has some \(j\in I\) such that \(x_i\in U\) for all \(i\ge j\).

Here is our main result on nets.

Proposition 1. Fix a subset \(A\) of a topological space \(X\). Then \(x\in\overline A\) if and only if there is a net of elements in \(A\) converging to \(x\).

Proof. Let’s begin with the backwards direction because it is more intuitive. Suppose that \(x\notin\overline A\), and we show that there is no net \(\{a_i\}_{i\in I}\) of elements in \(A\) converging to \(x\). Indeed, because \(x\notin\overline A\), we see that \(U=X\setminus\overline A\) is an open neighborhood of \(x\) disjoint from \(A\), so it has no intersection with no net of elements in \(A\).

The reverse direction is trickier because one has to construct a net. Indeed, suppose that \(x\in\overline A\). For our net, the trick is to choose the index set to be the collection \(I\) of open neighborhoods of \(x\), which we order by (reverse) inclusion. Now, for each open neighborhood \(U\) of \(x\), we need to produce an element of \(A\) which is close to \(x\); well, \(x\in\overline A\) implies that \(U\) cannot be disjoint with \(A\), so we simply take \(x_U\) to be any point in \(U\cap A\).

It remains to show that the net \(\{x_U\}\) converges to \(x\). Well, for any open neighborhood \(U\) of \(x\), we use the element \(U\in I\): for any \(U\le V\), we have \(V\subseteq U\), so \(x_V\in V\) is in \(U\) by construction. \(\blacksquare\)

And here is our main application.

Corollary 2. Let \(f\colon X\to Y\) be a function of topological spaces. Then \(f\) is continuous if and only if any convergent net \(\{x_i\}\to x\) goes to a convergent net \(\{f(x_i)\}\to f(x)\).

Proof. We show the directions independently.

• Suppose that \(f\) is continuous and that we are given a convergent net \(\{x_i\}\to x\). We need to show that the net \(\{f(x_i)\}\) converges to \(f(x)\). Well, for any open neighborhood \(V\) of \(f(x)\), we see that \(f^{-1}(V)\) is an open neighborhood of \(x\), so there is \(j\in I\) such that \(x_i\in f^{-1}(V)\) for all \(i\ge j\), which implies that \(f(x_i)\in V\) for all \(i\ge j\), as desired.

• Suppose that \(f\) sends convergent nets to convergent nets. For a closed subset \(B\subseteq Y\), we must show that \(f^{-1}(B)\subseteq X\) is closed. Well, being closed is equivalent to being equal to its closure, so choose some \(x\in\overline{f^{-1}(B)}\), and we will show that \(x\in f^{-1}(B)\).

  For this, we use the Proposition, which provides us with a net \(\{x_i\}\) from \(f^{-1}(B)\) such that \(\{x_i\}\to x\). Thus, \(\{f(x_i)\}\to f(x)\), so \(f(x)\in\overline B\) by the Proposition, so \(f(x)\in B\) because \(B\) is closed. We conclude \(x\in f^{-1}(B)\). \(\blacksquare\)

Approximately speaking, this tells us that nets are able to characterize the topology.

Corollary 3. Let \(f\colon X\to Y\) be a bijection of topological spaces. Then \(f\) is a homeomorphism if and only if each net \(\{x_i\}\) of \(X\) converges to \(x\) if and only if \(\{f(x_i)\}\) converges \(f(x)\).

Proof. The convergence of these nets is equivalent to \(f\) and \(f^{-1}\) both being continuous. \(\blacksquare\)

In other words, a topology on \(X\) is determined by its convergent nets.

It is also worthwhile to write down how the topology generated by a collection of functions behaves with respect to nets.

Proposition 4. Let \(X\) be a topological space whose topology is the weakest one for which a given collection of functions \(f_\alpha\colon X\to Y_\alpha\) are continuous. Then a net \(\{x_i\}\) in \(X\) converges to some \(x\) if and only if \(\{f_\alpha(x_i)\}\to f_\alpha(x)\) for each \(\alpha\).

Proof. In one direction, note the continuity of the maps \(f_\alpha\) implies that \(\{f_\alpha(x_i)\}\to f_\alpha(x)\) by Proposition 1.

In the reverse direction, suppose that \(\{f_\alpha(x_i)\}\to f_\alpha(x)\) for each \(\alpha\). To show that \(\{x_i\}\to x\), we want to show that the net is eventually in any open neighborhood \(U\) of \(x\). We may as well choose \(x\) to be a basic open neighborhood \(U\) of the form \[\bigcap_\alpha f_\alpha^{-1}(V_\alpha),\] where \(V_\alpha=Y_\alpha\) all but finitely often. Let \(I\) be a finite set of indices containing each \(\alpha\) such that \(V_\alpha\ne Y_\alpha\), so \[U=\bigcap_{\alpha\in I}f_\alpha^{-1}(V_\alpha).\] Now, because \(\{f_\alpha(x_i)\}\to f_\alpha(x)\) for each \(\alpha\), we are granted an index \(j_\alpha\) so that \(f_\alpha(x_i)\in V_\alpha\) for each \(i>j_\alpha\). Thus, we let \(j\) be any element exceeding \(\{j_\alpha:\alpha\in I\}\), from which we find that \(x_i\in U\) for \(i>j\). \(\blacksquare\)