From Neighborhoods to Nets
In a metric space, a sequence can often describe how points of a set approach a point in its closure. But a general topological space need not have a countable neighborhood base, so a sequence may not be enough. Nets provide a broader tool: their indices can record all the neighborhoods that need to be tested. The key construction assigns a point of the set to each neighborhood of the proposed limit.
Throughout, \((X,\mathcal{T})\) is a topological space. A directed set is a nonempty partially ordered set \((D,\leq)\) such that for any \(d,e\in D\), some \(f\in D\) satisfies \(d\leq f\) and \(e\leq f\). A net in \(X\) is a function \(d\mapsto x_d\) from a directed set \(D\) to \(X\). It converges to \(x\in X\) if, for every open neighborhood \(U\) of \(x\), there is \(d_0\in D\) such that \(x_d\in U\) whenever \(d\geq d_0\).
The neighborhood test works in every topological space. Nets will turn its condition—meeting \(A\) in every neighborhood—into one coherent family of points of \(A\). The direction from a convergent net to closure membership is immediate from the test; the reverse direction is the construction that makes nets especially useful.
A Limit of a Net in \(A\) Lies in \(\overline{A}\)
Proof. Let \((a_d)_{d\in D}\) be a net with \(a_d\in A\) for every \(d\), and suppose \(a_d\to x\). Let \(U\) be any open neighborhood of \(x\). By convergence, there is \(d_0\in D\) such that \(a_d\in U\) for every \(d\geq d_0\). In particular, \(a_{d_0}\in U\). Since \(a_{d_0}\in A\), we have \(U\cap A\ne\varnothing\). Every open neighborhood of \(x\) therefore intersects \(A\), so the neighborhood test gives \(x\in\overline{A}\). \(\square\)
This argument does not require \(x\notin A\), or that the terms of the net be distinct. If \(x\in A\), for example, the constant net with every term equal to \(x\) converges to \(x\).
Constructing a Net from the Neighborhoods
Proof. The proposition proves the implication from the existence of a convergent net to \(x\in\overline{A}\). For the converse, suppose \(x\in\overline{A}\). Let \(D\) be the set of all open neighborhoods of \(x\), ordered by reverse inclusion: \(U\leq V\) means \(V\subseteq U\). This is a nonempty directed set. It is nonempty because \(X\) is an open neighborhood of \(x\). If \(U,V\in D\), then \(U\cap V\) is also an open neighborhood of \(x\), and \(U\leq U\cap V\) and \(V\leq U\cap V\).
For every \(U\in D\), the neighborhood test gives \(U\cap A\ne\varnothing\). Choose \(a_U\in U\cap A\) for each \(U\in D\); this simultaneous selection uses the axiom of choice for the family of nonempty sets \(\{U\cap A:U\in D\}\). The assignment \(U\mapsto a_U\) is a net in \(A\). To check that it converges to \(x\), let \(W\) be any open neighborhood of \(x\). Use \(W\) itself as an index. If \(V\geq W\), the reverse-inclusion order gives \(V\subseteq W\), and hence \(a_V\in V\subseteq W\). Thus the net is eventually in every open neighborhood of \(x\), which is exactly convergence to \(x\). \(\square\)
The indexing order is essential: moving forward in the directed set means passing to a smaller neighborhood. As the index advances, the selected terms are forced into increasingly restrictive neighborhoods. The theorem applies to every topological space, whether or not it is first countable.
Worked Example: Approaching 0 from an Interval
Work in \(\mathbb{R}\) with its standard topology. Let \(A=(0,1)\) and \(x=0\). We construct an explicit net in \(A\) that converges to \(0\). Let \(D=(0,\infty)\), ordered by reverse numerical order: \(\varepsilon\leq\delta\) in \(D\) means \(\delta\leq\varepsilon\) as real numbers. This is directed, since \(\min(\varepsilon,\delta)\) is an upper bound for \(\varepsilon\) and \(\delta\) in this order.
For each \(\varepsilon\in D\), set \(a_\varepsilon=\min(\varepsilon/2,1/2)\). Both entries of the minimum are positive, so \(a_\varepsilon>0\), and \(a_\varepsilon\leq 1/2<1\). Thus every term lies in \(A\).
To verify convergence, take any open neighborhood \(W\) of \(0\). Since \(\mathbb{R}\) has its standard topology, there is \(r>0\) with \((-r,r)\subseteq W\). Choose an index \(\varepsilon_0\) satisfying \(0<\varepsilon_0<2r\). If \(\varepsilon\geq\varepsilon_0\) in the reverse order, then \(0<\varepsilon\leq\varepsilon_0\), so \(0<a_\varepsilon\leq\varepsilon/2\leq\varepsilon_0/2<r\). Therefore \(a_\varepsilon\in(-r,r)\subseteq W\), proving \(a_\varepsilon\to0\).
Check: for the neighborhood \(W=(-1/4,1/4)\), take \(r=1/4\) and \(\varepsilon_0=1/4\). Every later index has \(0<\varepsilon\leq1/4\), so \(0<a_\varepsilon\leq1/8<1/4\). This confirms eventual membership in \(W\).
Worked Example: A Net in the Sierpinski Space
Let \(X=\{a,b\}\) have the topology \(\mathcal{T}=\{\varnothing,\{a\},X\}\), and take \(A=\{a\}\). The open neighborhoods of \(b\) consist only of \(X\). In particular, every open neighborhood of \(b\) intersects \(A\), so the neighborhood test gives \(b\in\overline{A}\).
The set \(D\) of open neighborhoods of \(b\) has just one element, \(X\). Ordered by reverse inclusion, it is directed. The only possible selected term is \(a_X=a\), which lies in \(X\cap A\). The resulting net has one index and value \(a\). It converges to \(b\): the only open neighborhood of \(b\) is \(X\), and its sole net term belongs to \(X\).
This calculation also checks the theorem in a non-Hausdorff space, where one net can have distinct limits. Here the constant net at \(a\) converges to \(b\), even though its value is always \(a\), because convergence is determined by neighborhoods, not by a distance.
Why Sequences Can Be Insufficient
Worked Example: Closure in a Countable-Complement Topology
Let \(X\) be an uncountable set with the countable-complement topology: its open sets are \(\varnothing\) and all subsets whose complements in \(X\) are countable. Fix \(p\in X\) and set \(A=X\setminus\{p\}\). We show that \(p\in\overline{A}\) and construct a net in \(A\) converging to \(p\).
If \(U\) is an open neighborhood of \(p\), then \(U\ne\varnothing\) and \(X\setminus U\) is countable. If \(U\cap A\) were empty, then \(U\subseteq\{p\}\). Because \(p\in U\), this would give \(U=\{p\}\), whose complement \(X\setminus\{p\}\) is uncountable. That contradicts the definition of the topology. Hence every such \(U\) meets \(A\), and \(p\in\overline{A}\).
Now let \(D\) be the set of open neighborhoods of \(p\), ordered by reverse inclusion. For each \(U\in D\), choose \(a_U\in U\cap A\), which is possible by the preceding argument. If \(W\) is any open neighborhood of \(p\), then every index \(V\geq W\) satisfies \(V\subseteq W\), and consequently \(a_V\in W\). Thus \((a_U)_{U\in D}\) is a net in \(A\) converging to \(p\), as the theorem predicts.
In contrast, no sequence in \(A\) converges to \(p\). Given any sequence \((a_n)\) in \(A\), its range \(R=\{a_n:n\in\mathbb{N}\}\) is countable. Therefore \(U=X\setminus R\) is open and contains \(p\), but none of the sequence’s terms belongs to \(U\). The sequence is not eventually in this neighborhood. This verifies directly that sequences can fail to detect a closure point, while the net construction still works.
A Common Pitfall and the Scope of the Criterion
Common pitfall. One might argue, “Every neighborhood of \(p\) meets \(A\), so choose a sequence \(a_n\in A\) that eventually enters each neighborhood of \(p\).” The conclusion does not follow: in the countable-complement example, for any proposed sequence the open neighborhood \(X\setminus\{a_n:n\in\mathbb{N}\}\) contains \(p\) and misses every term. The correction is to index by all neighborhoods, not just by the natural numbers. For the net \((a_U)\), once the index reaches \(W\), every later index \(V\) satisfies \(V\subseteq W\), so its selected term lies in \(W\).
The theorem has two useful safeguards. First, it concerns membership in the closure, not necessarily being a limit point: when \(x\in A\), a constant net at \(x\) suffices. Second, the reverse implication depends on using a directed set so that any two neighborhood requirements can be refined together. Intersections of finitely many open neighborhoods of \(x\) are still open neighborhoods of \(x\), which is precisely what makes the reverse-inclusion order directed.
| Question | What the criterion says | Applies in |
|---|---|---|
| Does a convergent net in \(A\) have its limit in \(\overline{A}\)? | Yes; every neighborhood of the limit eventually contains a term of \(A\). | Every topological space |
| If \(x\in\overline{A}\), is there a net in \(A\) converging to \(x\)? | Yes; select a point from each neighborhood’s intersection with \(A\). | Every topological space |
| Can a sequence always replace the net? | No; a countable-complement topology can have closure points no sequence from \(A\) converges to. | Not in arbitrary spaces |
Check Your Understanding
Use the neighborhood definition of convergence and the closure test in your answers.
- Explain why a constant net at \(x\) converges to \(x\), and state when it is a net in \(A\).
- In the closure construction, why is the set of open neighborhoods of \(x\), ordered by reverse inclusion, directed?
- Suppose a net in \(A\) converges to \(x\). Show that an arbitrary open neighborhood of \(x\) intersects \(A\).
- In the countable-complement example, why does the complement of the range of any sequence give an open neighborhood of \(p\)?
- In the standard topology on \(\mathbb{R}\), verify directly that the net \(a_\varepsilon=\min(\varepsilon/2,1/2)\), indexed by positive \(\varepsilon\) in reverse order, converges to \(0\).