Functions as Coordinate Tuples
A function \(f:X\to Y\) can be recorded by listing its values \(f(x)\), one for each \(x\in X\). This identifies the set of all such functions with a product of copies of \(Y\), one copy for each domain point. Giving that product its product topology gives a topology on the function space: the pointwise topology. Its defining feature is that neighborhoods can control finitely many function values at a time, and its convergent sequences are exactly those that converge at every point of the domain.
Here \(X\) is a set and \((Y,\mathcal{T}_Y)\) is a topological space. No topology on \(X\) is needed for this construction. Write \(Y^X\) for the set of functions from \(X\) to \(Y\). The product \(\prod_{x\in X}Y\) consists of families \((y_x)_{x\in X}\) with \(y_x\in Y\); each factor is a copy of \(Y\). Its product topology has a basis consisting of sets \(\prod_{x\in X}U_x\), where every \(U_x\) is open in \(Y\) and \(U_x=Y\) except for finitely many indices.
A topology generated by a collection of sets consists of arbitrary unions of finite intersections of those sets. Thus each basic restriction in the pointwise topology concerns only finitely many values of a function. This finite-coordinate feature is precisely the one built into the product topology.
The Canonical Identification
Proof. First, \(\Phi\) is a bijection. A function \(f\) determines exactly one family of values \((f(x))_{x\in X}\), and a family \((y_x)_{x\in X}\) determines exactly one function by the rule \(f(x)=y_x\). If \(X=\varnothing\), there is one function with empty domain and one empty family, so the same assertion holds. If \(X\ne\varnothing\) and \(Y=\varnothing\), both sets are empty.
It remains to compare the topologies. Let \(\prod_{x\in X}U_x\) be a basic product open set, with \(U_x=Y\) except at finitely many indices. Its preimage under \(\Phi\) is \(\{f:f(x)\in U_x\text{ for every }x\in X\}\), which is the finite intersection of the sets \(\operatorname{ev}_x^{-1}(U_x)\) for which \(U_x\ne Y\). It is open in the pointwise topology. Conversely, each generating set \(\operatorname{ev}_x^{-1}(U)\) is the preimage under \(\Phi\) of the product open set that restricts coordinate \(x\) to \(U\) and leaves every other coordinate unrestricted. These two facts show that the topology transported by \(\Phi\) is exactly the product topology. Since \(\Phi\) is a bijection and both it and its inverse preserve open sets, it is a homeomorphism. When \(X=\varnothing\), both spaces are singletons and have their unique topology, also as asserted. \(\square\)
The theorem gives a concrete way to read the pointwise topology: it is not an additional structure unrelated to products. It is the product topology written in the language of functions, with the evaluation maps serving as coordinate projections.
Worked Example: Two Coordinates and a Convergent Sequence
Let \(X=\{a,b\}\) and let \(Y=\mathbb{R}\) with its standard topology. Define \(f_n\in\mathbb{R}^X\), for \(n\in\mathbb{N}\), by \(f_n(a)=1/n\) and \(f_n(b)=2-1/n\). The canonical identification represents \(f_n\) by the pair \((1/n,2-1/n)\).
At coordinate \(a\), the difference from \(0\) is \(|f_n(a)-0|=|1/n|=1/n\), which tends to \(0\). At coordinate \(b\), the difference from \(2\) is \(|f_n(b)-2|=|(2-1/n)-2|=1/n\), which also tends to \(0\). Therefore \(f_n\) converges pointwise to the function \(f\) with \(f(a)=0\) and \(f(b)=2\). In the product representation, this is the convergence \((1/n,2-1/n)\to(0,2)\).
Check: For \(n=4\), the pair is \((1/4,7/4)\). Its coordinate differences from \((0,2)\) are \(|1/4-0|=1/4\) and \(|7/4-2|=1/4\), agreeing with the formula \(1/n\) at \(n=4\).
Continuity into a Pointwise Function Space
Proof. If \(g\) is continuous, then each \(\operatorname{ev}_x\circ g\) is continuous because \(\operatorname{ev}_x\) is continuous by the definition of the pointwise topology, and a composition of continuous maps is continuous. For the reverse implication, suppose every \(\operatorname{ev}_x\circ g\) is continuous. A generating open set in \(Y^X\) has the form \(\operatorname{ev}_x^{-1}(U)\), where \(U\) is open in \(Y\). Its preimage under \(g\) is \((\operatorname{ev}_x\circ g)^{-1}(U)\), which is open in \(Z\). Since preimages preserve unions and finite intersections, preimages of all open sets generated by these sets are open. Hence \(g\) is continuous. If \(X\) is empty, the target \(Y^X\) is a singleton; every map into it is continuous, and the coordinate condition is vacuous, so the equivalence still holds. \(\square\)
Worked Example: A Continuous Family of Functions
Let \(X=\{u,v\}\), \(Y=\mathbb{R}\), and \(Z=\mathbb{R}\), all with their standard topologies. For \(t\in\mathbb{R}\), define \(g(t)\in\mathbb{R}^X\) by \(g(t)(u)=2t+1\) and \(g(t)(v)=t^2\). The two coordinate maps are \(t\mapsto 2t+1\) and \(t\mapsto t^2\).
To verify their continuity, fix \(t_0\in\mathbb{R}\) and \(\varepsilon>0\). For the first coordinate, if \(|t-t_0|<\varepsilon/2\), then \(|(2t+1)-(2t_0+1)|=2|t-t_0|<\varepsilon\). For the second, first require \(|t-t_0|<1\), which gives \(|t+t_0|\le 2|t_0|+1\). Taking \(\delta=\min\{1,\varepsilon/(2|t_0|+1)\}\), the condition \(|t-t_0|<\delta\) implies \(|t^2-t_0^2|=|t-t_0||t+t_0|<\varepsilon\). Thus both coordinate maps are continuous, and Theorem 2 implies that \(g\) is continuous.
For a direct value check, \(g(2)(u)=2(2)+1=5\) and \(g(2)(v)=2^2=4\), so its product coordinates are \((5,4)\). These values agree with the defining formulas.
Convergence Is Coordinatewise
In any topological space, a sequence converges to a point if it is eventually in every open neighborhood of that point. In \(Y^X\), pointwise convergence means that for each fixed \(x\in X\), the sequence of values in \(Y\) converges to the corresponding value of the limiting function.
Proof. Suppose first that \(f_n\to f\) in \(Y^X\). For each \(x\in X\), the evaluation map \(\operatorname{ev}_x\) is continuous. Therefore \(\operatorname{ev}_x(f_n)\to\operatorname{ev}_x(f)\), which says exactly that \(f_n(x)\to f(x)\).
Conversely, suppose \(f_n(x)\to f(x)\) for every \(x\in X\). Take any basic open neighborhood of \(f\) in the pointwise topology. It imposes conditions \(f_n(x)\in U_x\) at a finite set \(F\subseteq X\), where each \(U_x\) is open and contains \(f(x)\). For each \(x\in F\), coordinate convergence gives an integer \(N_x\) such that \(f_n(x)\in U_x\) whenever \(n\ge N_x\). If \(F\ne\varnothing\), let \(N\) be the largest of these finitely many integers. Then for every \(n\ge N\), all the finitely many required conditions hold at once, so \(f_n\) is in the chosen neighborhood. If \(F=\varnothing\), the neighborhood imposes no restrictions and contains every \(f_n\). Thus every basic neighborhood of \(f\) eventually contains the sequence, proving \(f_n\to f\). \(\square\)
Worked Example: Functions Supported at One Index
Take \(X=\mathbb{N}\) and \(Y=\mathbb{R}\) with the standard topology. For each \(n\in\mathbb{N}\), define \(f_n(k)=1\) when \(k=n\), and \(f_n(k)=0\) when \(k\ne n\). Let \(f(k)=0\) for every \(k\in\mathbb{N}\).
Fix any coordinate \(k\). If \(n>k\), then \(n\ne k\), so \(f_n(k)=0=f(k)\). In particular, the coordinate sequence is eventually equal to its limit, and therefore converges to it in \(\mathbb{R}\). Since this holds for each \(k\), Theorem 3 gives \(f_n\to f\) in the pointwise topology.
Check using a basic neighborhood: Suppose a neighborhood restricts the coordinates in a finite set \(F\subseteq\mathbb{N}\). If \(F\ne\varnothing\), choose an integer \(N\) larger than every element of \(F\). For \(n\ge N\) and \(k\in F\), we have \(n\ne k\), hence \(f_n(k)=0=f(k)\), so every required coordinate condition holds. If \(F=\varnothing\), there are no conditions to check. This verifies convergence directly from the product basis.
A Common Pitfall: Restricting Every Coordinate
A common error is to treat a product neighborhood as if it had to restrict every coordinate. The product topology does not work that way: a basic open set can restrict only finitely many coordinates, leaving all the others unrestricted. Requiring restrictions at every coordinate instead describes the box topology, which can be strictly finer than the product topology.
For a concrete distinction, take \(\mathbb{R}^{\mathbb{N}}\) with the product topology and the set \(B=\prod_{k\in\mathbb{N}}(-1/2,1/2)\). The set contains the zero sequence, but it is not product-open. Indeed, any basic product neighborhood of the zero sequence restricts only a finite set \(F\) of coordinates. Choose \(k\notin F\); the sequence that is \(2\) at coordinate \(k\) and \(0\) elsewhere belongs to that basic neighborhood, because its coordinates in \(F\) are zero, but it does not belong to \(B\), because \(2\notin(-1/2,1/2)\). Thus no basic product neighborhood of zero is contained in \(B\).
The earlier sequence \(f_n\) illustrates the consequence of the mistaken rule. Every \(f_n\) has one coordinate equal to \(1\), so none belongs to \(B\); demanding eventual membership in \(B\) would incorrectly rule out convergence to zero. In the product topology, a neighborhood tests only a finite set \(F\). Choosing \(n\) larger than every element of \(F\) makes \(f_n(k)=0\) at every tested coordinate, which is exactly why the sequence converges pointwise.
The central practical distinction is therefore between checking each coordinate separately and imposing a single condition on all coordinates at once. Product convergence is characterized by the first: each fixed coordinate must converge, and finite-coordinate neighborhoods ensure those separate conditions can be met together.
Check Your Understanding
Use the product basis and evaluation maps to justify your answers.
- For \(X=\{a,b\}\), write the canonical product coordinates of a function \(f:X\to\mathbb{R}\).
- Explain why the pointwise topology on \(Y^X\) is the coarsest topology making every evaluation map continuous.
- Let \(X=\mathbb{N}\) and \(f_n(k)=1\) if \(k\le n\), and \(f_n(k)=0\) otherwise. Find the pointwise limit and verify convergence at a fixed coordinate \(k\).
- Use Theorem 2 to show that \(t\mapsto g(t)\), where \(g(t)(u)=t\) and \(g(t)(v)=3t\), is continuous from \(\mathbb{R}\) to \(\mathbb{R}^{\{u,v\}}\).
- Why does a basic product neighborhood restrict only finitely many coordinates, and how does that fact enter the proof of the convergence theorem?