Nets in Uncountable Pointwise Function Spaces

Finite Coordinate Tests

In the pointwise topology, a neighborhood places conditions on only finitely many input points at a time. This finite-coordinate feature makes it possible to approximate any function using one with finite support. But on an uncountable domain, no sequence of finite supports can eventually reach every input point. Nets, whose index sets need not be countable, capture the resulting convergence.

Let \(X\) be a set, and give \(\mathbb{R}\) its standard topology. The function space \(\mathbb{R}^X\) is the set of all functions \(f:X\to\mathbb{R}\), with the pointwise topology, equivalently the product topology on \(\prod_{x\in X}\mathbb{R}\). A basic open neighborhood of \(f\in\mathbb{R}^X\) is specified by a finite set \(E\subseteq X\) and open sets \(V_x\subseteq\mathbb{R}\), one for each \(x\in E\), with \(f(x)\in V_x\). It consists of the functions \(h\) satisfying \(h(x)\in V_x\) for every \(x\in E\); there are no restrictions at points outside \(E\).

A net is a family indexed by a directed set, which can be much larger or less familiar than \(\mathbb{N}\). A directed set is a nonempty set \(J\) with an order \(\leq\) such that any two elements of \(J\) have a common upper bound. A net in a space \(Y\) is a function \(j\mapsto y_j\) from \(J\) to \(Y\). It converges to \(y\in Y\) if, for every open neighborhood \(U\) of \(y\), there is \(j_0\in J\) such that \(y_j\in U\) whenever \(j\geq j_0\). This is called being eventually in \(U\). A sequence is the special case of a net indexed by \(\mathbb{N}\) in its usual order.

Finitely Supported Functions Are Dense

Proof. A subset is dense if and only if every nonempty open set intersects it. To verify this criterion, if \(D\) is dense, every point of any nonempty open set \(U\) lies in \(\overline{D}\), so the neighborhood test for closure gives \(U\cap D\ne\varnothing\). Conversely, if every nonempty open set meets \(D\), no point can have an open neighborhood disjoint from \(D\), so the neighborhood test gives \(\overline{D}=\mathbb{R}^X\).

Now take any nonempty open set in \(\mathbb{R}^X\), choose a function \(f\) in it, and choose a basic open neighborhood of \(f\) contained in that set. Let \(E\subseteq X\) be the finite set of coordinates constrained by this basic neighborhood. Define \(g:X\to\mathbb{R}\) by \(g(x)=f(x)\) for \(x\in E\) and \(g(x)=0\) for \(x\notin E\). Then \(\operatorname{supp}(g)\subseteq E\), so \(g\in D\). At every constrained coordinate \(x\in E\), \(g(x)=f(x)\in V_x\). Thus \(g\) belongs to the chosen basic neighborhood and hence to the original open set. The argument also applies when \(E=\varnothing\): then \(g\) is the zero function and the basic neighborhood imposes no constraints. Therefore every nonempty open set intersects \(D\), proving density. \(\square\)

A Net That Converges to the Constant-One Function

Assume now that \(X\) is uncountable. Let \(\mathcal{F}(X)\) be the set of finite subsets of \(X\), ordered by inclusion. This is a directed set: it is nonempty because it contains \(\varnothing\), and any two finite subsets \(F,G\) have the common upper bound \(F\cup G\), which is finite. For each \(F\in\mathcal{F}(X)\), define \(g_F:X\to\mathbb{R}\) to equal \(1\) on \(F\) and \(0\) on \(X\setminus F\). Each \(g_F\) is finitely supported.

Proof. Let \(U\) be any basic open neighborhood of \(\mathbf{1}\). It constrains a finite set \(E\subseteq X\), and at each \(x\in E\) its permitted open set \(V_x\) contains \(1\). Choose the index \(F_0=E\). If \(F\geq F_0\) in the inclusion order, then \(E\subseteq F\), so \(g_F(x)=1=\mathbf{1}(x)\) for every \(x\in E\). In particular \(g_F(x)\in V_x\) at every constrained coordinate, and \(g_F\in U\). Thus the net is eventually in every basic neighborhood of \(\mathbf{1}\). Every open neighborhood contains a basic one, so the net converges to \(\mathbf{1}\). This proof includes \(E=\varnothing\), for which \(F_0=\varnothing\) and there are no coordinate restrictions to check. \(\square\)

Why No Sequence Can Do the Same

Proof. Suppose \((f_n)_{n\in\mathbb{N}}\) is any sequence of finitely supported functions, and put \(S_n=\operatorname{supp}(f_n)\). Each \(S_n\) is finite. The union \(S=\bigcup_{n\in\mathbb{N}}S_n\) is countable, since it is a countable union of finite sets. Because \(X\) is uncountable, choose \(x\in X\setminus S\). For every \(n\), \(x\notin S_n\), so the definition of support gives \(f_n(x)=0\). The open interval \((\frac12,\frac32)\) is a neighborhood of \(1\), but \(f_n(x)=0\notin(\frac12,\frac32)\) for every \(n\). Therefore the coordinate sequence \((f_n(x))\) does not converge to \(1\). A basic neighborhood of \(\mathbf{1}\) restricting just coordinate \(x\) is never entered by the sequence, so the sequence cannot converge to \(\mathbf{1}\) in the pointwise topology. \(\square\)

A Common Pitfall and the Role of Nets

Common pitfall. One might try to list the points of uncountable \(X\) as \(x_1,x_2,\ldots\) and let the \(n\)-th finite support contain the first \(n\) points. This assumes that the list covers \(X\). In fact, the union of the supports in any sequence is countable, so it cannot contain every point of an uncountable \(X\). The correction is to index by all finite subsets of \(X\), ordered by inclusion. For any finite group of coordinates that a neighborhood tests, there is an index containing all of them; no enumeration of \(X\) is needed.

The distinction is not that sequences fail to describe pointwise convergence in general: a sequence converging in the product topology must converge at each coordinate. Here the obstruction is that a sequence of finitely supported functions can involve only countably many coordinates across all its terms. A net indexed by finite subsets can eventually include each finite collection of coordinates, even though the whole domain is uncountable. This is why the density result and the net construction coexist with the failure of every such sequence to reach the constant-one function.