Directed Sets and Nets

Beyond Sequences

A sequence is a function indexed by \(\mathbb{N}\), so its terms have a familiar order: each term has a later term, and every two indices have a common later index. Topological spaces need not have countable neighborhood bases, however. To describe approach using all neighborhoods of a point, we need an index set that can have many directions while still allowing indices to move consistently “later.” Directed sets provide that structure, and functions from them are called nets.

Throughout, \((X,\mathcal{T})\) is a topological space. A neighborhood of \(x\in X\) means an open set containing \(x\). Recall the neighborhood test for closure: \(x\in\overline{A}\) if and only if every open neighborhood of \(x\) intersects \(A\). The constructions below use this test to choose net values from \(A\).

The order records which indices count as later, not an order on the values \(x_d\). The common-upper-bound condition ensures that two stages can be followed by a stage at least as late as both. It also extends to any finite collection of indices: repeatedly apply the condition to the common upper bound already obtained and the next index. No corresponding requirement says that every pair of indices must be comparable.

Sequences and More Flexible Index Sets

The natural numbers with their usual order form a directed set: it is nonempty, and \(\max(m,n)\) is an upper bound for \(m,n\). Thus every sequence \((x_n)_{n\in\mathbb{N}}\) is a net. A net is not required to be a sequence: its index set may contain incomparable indices or may be much larger than \(\mathbb{N}\).

Directed sets can also record several requirements independently. A product order is a standard way to do this. If \(D_1,D_2\) are directed sets, order \(D_1\times D_2\) coordinate by coordinate: \((a,b)\leq(c,d)\) means \(a\leq c\) and \(b\leq d\).

Proof. The product is nonempty because each factor is nonempty. The coordinatewise relation is a partial order: reflexivity, antisymmetry, and transitivity each follow from the corresponding property in both factors. Given \((a,b),(c,d)\in D_1\times D_2\), directedness of \(D_1\) gives \(e\in D_1\) with \(a\leq e\) and \(c\leq e\), while directedness of \(D_2\) gives \(f\in D_2\) with \(b\leq f\) and \(d\leq f\). Therefore \((e,f)\) is an upper bound of both product indices. \(\square\)

Indexing by Neighborhoods

Fix \(x\in X\), and let \(D_x\) be the set of all open neighborhoods of \(x\). Order it by reverse inclusion: \(U\leq V\) means \(V\subseteq U\). Thus a smaller neighborhood is a later index. This order is directed because \(U\cap V\) is an open neighborhood of \(x\) contained in both \(U\) and \(V\); consequently \(U\leq U\cap V\) and \(V\leq U\cap V\).

Proof. The set \(D_x\) is nonempty because \(X\) is an open neighborhood of \(x\). Reverse inclusion is a partial order: inclusion is reflexive and transitive, and if \(U\subseteq V\) and \(V\subseteq U\), then \(U=V\). For any \(U,V\in D_x\), their intersection is open and contains \(x\), so \(U\cap V\in D_x\). Since \(U\cap V\subseteq U\) and \(U\cap V\subseteq V\), the reverse-inclusion order gives \(U\leq U\cap V\) and \(V\leq U\cap V\). This is a common upper bound. \(\square\)

This directed set supports a general construction. Suppose \(A\subseteq X\) and \(x\in\overline{A}\). By the neighborhood test, \(A\cap U\neq\varnothing\) for every \(U\in D_x\). Choose one point \(a_U\in A\cap U\) for each \(U\), and define a net indexed by \(D_x\) with value \(a_U\) at \(U\). For an arbitrary family of nonempty sets \(A\cap U\), making all these choices uses the axiom of choice. In concrete examples, the choice can often be specified by a rule.

The construction records the approach directly: at the index \(U\), the selected value lies in \(U\). If an index \(V\) is later than \(U\), then \(V\subseteq U\), so its selected value lies in the smaller neighborhood \(V\). When \(x\in A\), this construction may select \(x\) itself. If the goal is to select values different from \(x\), the relevant hypothesis is that \(x\) is a limit point of \(A\): every neighborhood then contains a point of \(A\setminus\{x\}\).

The same neighborhood-indexing idea works when no countable list of neighborhoods is available. For instance, let \(X=\mathbb{R}\) have the cocountable topology: its open sets are \(\varnothing\) and the sets whose complements are countable. Fix \(p\in X\) and put \(A=X\setminus\{p\}\). Every open neighborhood \(U\) of \(p\) has countable complement, so it is uncountable and cannot be contained in \(\{p\}\). Hence \(U\cap A\neq\varnothing\). Choosing \(a_U\in U\cap A\) for every such \(U\) gives a net with values in \(A\), indexed by all open neighborhoods of \(p\). This construction does not require the neighborhoods to be listed in a sequence.

A Common Ordering Mistake

Common pitfall. Using ordinary inclusion for neighborhoods reverses which indices count as later. For example, let \(U_1=(-1,1)\) and \(U_2=(-1/2,1/2)\). Since \(U_2\subset U_1\), the reverse-inclusion order gives \(U_1\leq U_2\): the smaller neighborhood is later, as intended. If ordinary inclusion is used instead, then \(U_2\leq U_1\), so the larger neighborhood is later. In the family \(U_n=(-1/n,1/n)\), ordinary inclusion makes \(U_1\) an upper bound of every \(U_n\), rather than directing indices toward smaller neighborhoods. The correction is to order neighborhoods by reverse inclusion.

Sequences are useful whenever countably many indices capture the required stages. Nets broaden the available indexing structure: product orders can track separate requirements, and the full neighborhood system can encode approach even without a countable neighborhood base. Their value here is the flexibility of the index set; the precise relationship between net convergence and neighborhood tests is a separate topic.