The topology of pointwise convergence on $Y^X$, which is the product topology, and its restriction to $C(X,Y)$

Let $X$ be a set and let $(Y,{\mathcal{T}}_Y)$ be a topological space (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison). Write

$Y^X:={\prod }_{x\in X}Y,$

the product of the constant family whose factor at every index $x\in X$ is $Y$ (The product set ${\prod }_{i\in I}{X}_i$ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space). Unwinding that definition, an element of $Y^X$ is a function with domain $X$ taking its value at each $x$ in $Y$; so $Y^X$ is the set of all functions $X\to Y$, and the projection at $x$ is evaluation,

${\pi }_x:Y^X\to Y,{\pi }_x(f)=f(x)\mathrm{.}$

The topology of pointwise convergence on $Y^X$ is the product topology: the initial topology of the family $({\pi }_x{)}_{x\in X}$ (The initial topology of a family of maps into spaces and the final topology of a family of maps out of spaces, and the subspace topology as the model initial topology), that is the topology generated by the subbasis

$\mathcal{G}:=\{{\pi }_x^{-1}[V]:x\in X,V\in {\mathcal{T}}_Y\},{\pi }_x^{-1}[V]=\{f\in Y^X:f(x)\in V\}\mathrm{.}$

By A family is a basis for a unique topology iff it covers the set and every point of an intersection of two members lies in a member inside that intersection; finite intersections of any subbasis form a basis the finite intersections of members of $\mathcal{G}$ form a basis for it (Basis and subbasis for a topology, and the topology generated by a family of sets), so the basic open sets are exactly the sets

$\{f\in Y^X:f(x_j)\in V_j\text{ for every }j<n\}(n\in \mathbb{N},x_0,\dots ,x_{n-1}\in X,V_0,\dots ,V_{n-1}\in {\mathcal{T}}_Y),$

the value $n=0$ giving the empty intersection $Y^X$ itself. A basic open set therefore constrains a member of $Y^X$ at finitely many points only, and that is the whole content of the topology.

The restriction to the continuous maps. Suppose in addition that $X$ carries a topology, and write

$C(X,Y):=\{f\in Y^X:f\text{ is continuous}\}$

(Continuity of a map of topological spaces at a point and globally). The topology of pointwise convergence on $C(X,Y)$ is the subspace topology inherited from $Y^X$ (Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace); its subbasic open sets are the traces ${\pi }_x^{-1}[V]\cap C(X,Y)$, since tracing carries a subbasis to a subbasis (Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace).

Nothing on this page gives $C(X,Y)$ a default topology. The set $C(X,Y)$ carries several different topologies below, and every statement names the one it means at the point of use.

• Why the name. A sequence converges in this topology exactly when it converges at every point of $X$; that is the next item, and it is what justifies calling the product topology on a set of functions the topology of pointwise convergence. Nothing about sequences is built into the definition, which is purely the product topology of The product set ${\prod }_{i\in I}{X}_i$ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space.

• The box topology is a different topology on the same set, and a finer one (The product set ${\prod }_{i\in I}{X}_i$ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space). It is not used here: the topology of pointwise convergence is the product topology, and the product topology is what has the characteristic property of A map into a product is continuous iff each of its components is; the projections are continuous and open; and each projection is surjective when every factor is nonempty, which for an infinite index set uses the Axiom of Choice.

• $Y^X$ here carries no algebra. The vector space $F^X$ of all functions $X\to F$ with pointwise operations, and $F^n$ as the case $X=n=\{0,1,\dots ,n-1\}$ writes $F^X$ for the same set of functions when the target is a field, and equips it with pointwise addition and scalar multiplication to make a vector space. That is a different structure on the same underlying set: nothing on this page uses those operations, and nothing on this page requires the target to be a field or even to have a single algebraic operation. Where both are in play the algebraic structure is named explicitly.

• Degenerate cases are not excluded. For $X=\mathrm{\varnothing }$ the set $Y^X$ has exactly one element, the empty function, and carries its unique topology (The product set ${\prod }_{i\in I}{X}_i$ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space). For $Y$ empty and $X$ nonempty, $Y^X$ is empty. Both are consistent with the definition and neither is used below; the items on this page that need $X$ nonempty say so.