Nets and filters repair the defects of sequences in non-first-countable spaces. They give exact convergence tests for closure, continuity, and compactness.
Nets and directed sets
A net is a sequence indexed by a directed set rather than by the natural numbers.
Definition. A directed set is a preorder \(D\) such that for any \(d_1,d_2\in D\) there exists \(d_3\in D\) with \(d_1\le d_3\) and \(d_2\le d_3\). A net in \(X\) is a function \(x:D\to X\), usually written \((x_d)_{d\in D}\). It converges to \(x\) if for every neighborhood \(U\) of \(x\), there is \(d_0\) such that \(d\ge d_0\) implies \(x_d\in U\).
Example. Sequences are nets indexed by \(\mathbb N\). Neighborhoods of a point ordered by reverse inclusion also form a directed set, which is the canonical index set for closure arguments.
Closure by nets
Nets give an exact replacement for the sequence closure criterion.
Theorem. For \(A\subseteq X\), a point \(x\) belongs to \(\overline A\) if and only if there is a net in \(A\) converging to \(x\).
Proof. If a net in \(A\) converges to \(x\), then every neighborhood of \(x\) eventually contains terms of the net and therefore meets \(A\), so \(x\in\overline A\). Conversely, if \(x\in\overline A\), let \(D\) be the set of neighborhoods of \(x\), ordered by reverse inclusion. For each \(U\in D\), choose \(x_U\in A\cap U\). Then \((x_U)\) is a net in \(A\) converging to \(x\).
Corollary. A map \(f:X\to Y\) is continuous if and only if \(x_d\to x\) implies \(f(x_d)\to f(x)\) for every net.
Proof. Continuity immediately preserves convergence. Conversely, if \(f\) is not continuous, there is \(A\subseteq X\) with \(f(\overline A)\not\subseteq\overline{f(A)}\). Choose \(x\in\overline A\) witnessing this, take a net in \(A\) converging to \(x\), and use the assumed net preservation to get a contradiction.
Filters and convergence
Filters encode eventual membership without choosing indexed points.
Definition. A filter \(\mathcal F\) on \(X\) is a nonempty family of subsets of \(X\) closed upward under supersets, closed under finite intersections, and not containing \(\varnothing\). It converges to \(x\) if every neighborhood of \(x\) belongs to \(\mathcal F\).
Proposition. Every net determines an eventuality filter, and every filter determines a net whose convergence behavior matches the filter.
Proof. For a net \((x_d)\), take the filter generated by tails \(\{x_e:e\ge d\}\). If the net converges to \(x\), every neighborhood contains a tail, hence lies in the filter. Conversely, given a filter \(\mathcal F\), index pairs \((F,x)\) with \(F\in\mathcal F\) and \(x\in F\), ordered by reverse inclusion of \(F\); the resulting net has tails inside each filter set.
Compactness by ultrafilters
The most compact convergence criterion uses maximal filters.
Definition. An ultrafilter is a filter maximal under inclusion. Equivalently, for every \(A\subseteq X\), exactly one of \(A\) and \(X\setminus A\) belongs to the ultrafilter.
Theorem. A space \(X\) is compact if and only if every ultrafilter on \(X\) converges to at least one point.
Proof. If \(X\) is compact and \(\mathcal U\) is an ultrafilter with no limit, then for every \(x\in X\) choose a neighborhood \(V_x\notin\mathcal U\). The \(V_x\) cover \(X\), so finitely many cover \(X\). Their complements all lie in \(\mathcal U\), so their finite intersection lies in \(\mathcal U\), but that intersection is empty, impossible. Conversely, if \(X\) is not compact, choose a closed family with the finite intersection property and extend it to an ultrafilter. Any limit point would belong to every closed set in the family, contradicting empty total intersection.