Using Nets to Track Uncountable Neighborhood Behavior

Why Use Nets?

In a topological space, neighborhoods of a point may not be organized by a countable list. A sequence has only countably many terms, so a sequence-based test may not record the behavior of every neighborhood. A net generalizes a sequence by allowing its terms to be indexed by a more flexible ordered set. The key idea is to index directly by neighborhoods: as the index advances, the neighborhoods become smaller.

Throughout, \((X,\mathcal{T})\) and \((Y,\mathcal{U})\) are topological spaces. A directed set is an index set with an order that allows any two indices to be followed by a common later index. We use a reflexive, transitive order; distinct indices are allowed to be comparable in both directions.

The phrase “whenever \(i_0\preceq i\)” is the eventual condition: every term at or after \(i_0\) must lie in \(U\). A sequence is a net whose directed set is \(\mathbb{N}\) with its usual order. The extra flexibility of nets comes from the choice of directed set, not from changing the definition of convergence.

Directing the Neighborhoods

Let \(x\in X\). The collection of all open neighborhoods of \(x\) becomes directed when ordered by reverse inclusion: define \(U\preceq V\) when \(V\subseteq U\). Thus advancing in the order means passing to a smaller neighborhood. For any two neighborhoods \(U,V\), their intersection is another open neighborhood of \(x\), and it is a common later index.

Sometimes it is useful to attach a point of a set to each neighborhood. To avoid making an arbitrary selection for every neighborhood, use indices that are pairs consisting of a neighborhood and one of its points. The following result uses this construction to turn the neighborhood test for closure into a convergent net.

Proof. Suppose first that \(x\in\overline{A}\). The neighborhood test for closure says that every open neighborhood \(U\) of \(x\) intersects \(A\). Define \(D=\{(U,a): U\text{ is an open neighborhood of }x,\ a\in U\cap A\}\). This set is nonempty. Put \((U,a)\preceq(V,b)\) when \(V\subseteq U\). Given \((U,a),(V,b)\in D\), the set \(U\cap V\) is an open neighborhood of \(x\), so it meets \(A\); choose \(c\in U\cap V\cap A\). Then \((U\cap V,c)\) is a common later index, so \(D\) is directed. Define \(a_{(U,a)}=a\). Every term belongs to \(A\). If \(W\) is any open neighborhood of \(x\), take an index \((W,w)\in D\). At every later index \((V,b)\), we have \(V\subseteq W\) and \(b\in V\), so \(b\in W\). Thus \(a_i\to x\).

Conversely, suppose a net with all terms in \(A\) converges to \(x\). If some open neighborhood \(W\) of \(x\) missed \(A\), then no term of the net would lie in \(W\). This contradicts convergence, which requires all terms after some index to lie in \(W\). Hence every open neighborhood of \(x\) meets \(A\), and the neighborhood test gives \(x\in\overline{A}\). If \(A=\varnothing\), its closure is empty and there is no net with values in \(A\), so the equivalence also covers this case. \(\square\)

Neighborhood-Indexed Nets in a Large Topology

For the neighborhood-directed construction, the index set need not be countable. This matters in spaces whose neighborhoods are specified by excluding many points at once.

Continuity Can Also Be Tested by Nets

Proof. Suppose \(f\) is continuous, let \(x_i\to x\), and let \(V\) be an open neighborhood of \(f(x)\). By continuity, \(f^{-1}(V)\) is open and contains \(x\). Convergence gives an index \(i_0\) such that \(x_i\in f^{-1}(V)\) for all \(i\succeq i_0\). Thus \(f(x_i)\in V\) for all such \(i\), proving \(f(x_i)\to f(x)\).

Conversely, suppose \(f\) is not continuous. Then there is \(x\in X\) and an open neighborhood \(V\) of \(f(x)\) such that no open neighborhood \(U\) of \(x\) satisfies \(f(U)\subseteq V\). For each such \(U\), there is therefore \(y\in U\) with \(f(y)\notin V\). Let \(D=\{(U,y):U\text{ is an open neighborhood of }x,\ y\in U,\ f(y)\notin V\}\), ordered by reverse inclusion of the neighborhood coordinates. This set is directed: for two indices, \(U\cap W\) is a neighborhood of \(x\), and the assumed failure of continuity supplies a point \(z\in U\cap W\) with \(f(z)\notin V\). Define \(x_{(U,y)}=y\). For any neighborhood \(W\) of \(x\), all terms after an index with neighborhood coordinate \(W\) lie in \(W\), so \(x_{(U,y)}\to x\). But every image lies outside \(V\), so the image net cannot converge to \(f(x)\), since \(V\) is a neighborhood of \(f(x)\). This contradicts the assumed preservation of all convergent nets. Therefore \(f\) is continuous. \(\square\)

A Common Pitfall: Ordering Neighborhoods the Wrong Way

When neighborhoods themselves are the indices, they must be ordered so that later indices represent smaller neighborhoods. Reversing this order can invalidate the intended convergence argument.

Wrong calculation. In \(\mathbb{R}\), let \(U_r=(-r,r)\) for \(r>0\), and assign the value \(x_{U_r}=r/2\in U_r\). If the index order is inclusion, then \(U_r\preceq U_s\) when \(r\leq s\): the neighborhoods grow as the net advances. The index \(U_1\) is followed by \(U_4\), whose value is \(x_{U_4}=2\), outside the neighborhood \((-1,1)\) of \(0\). In fact, after any starting index \(U_r\), one can take \(s>\max(r,2)\); then \(U_r\preceq U_s\) but \(x_{U_s}=s/2>1\). The net does not converge to \(0\).

Correction. Order the same neighborhoods by reverse inclusion: \(U_r\preceq U_s\) when \(s\leq r\). Given \(\varepsilon>0\), choose \(r_0<2\varepsilon\). Every later \(U_s\) has \(s\leq r_0\), and hence \(0<x_{U_s}=s/2\leq r_0/2<\varepsilon\). Thus the net converges to \(0\). The order is part of the net, not a cosmetic choice.