Topologies on Function Spaces

Munkres Topology

Here we introduce four canonical topologies imposed on function spaces of the form $X^Y$, where $X$ and $Y$ are topological spaces. They are as follows: the uniform topology, the topology of compact convergence, the compact-open topology, and the pointwise convergence topology. Of these, the pointwise convergence and uniform topologies should be familiar from elementary analysis and topology, whereas the compact convergence and compact-open topology are relatively novel concepts. Furthermore, the uniform and compact convergence topologies require $Y$ to be a metric space to be defined, while the compact-open and pointwise convergence topologies can be defined with general spaces $X$ and $Y$. As a final notational note, we use $\mathcal{C}(X,Y)$ to denote the space of continuous functions from $X$ to $Y$.

We first enumerate their definitions, beginning with the compact-open and pointwise convergence topologies which don’t require $Y$ to have a metric. If $C$ is a compact subset of $X$ and $U$ is an open subset of $Y$, let

$$S(C,U)=\{f|f\in X^Y\text{ and }f(C)\subset U\}$$

The sets $S(C,U)$, as a subbasis, generates a topology on $X^Y$ which we call the compact-open topology. Unlike the other three topologies, we typically only use the compact-open topology as defined in $\mathcal{C}(X,Y)$ instead of the entire space $X^Y$. If $x$ is a point of $X$ and $U$ is an open subset of $Y$, let

$$S(x,U)=\{f|f\in X^Y\text{ and }f(x)\in U\}$$

The sets $S(x,U)$ as a subbasis generates the pointwise convergence topology in $X^Y$. Note this is precisely the familiar product topology as $S(x,U)={\pi }^{-1}(U)$ where $\pi$ is the projection from $X^Y$ onto the $x\in X$ coordinate. It is called the pointwise convergence topology as a sequence $f_n$ converges to $f$ in this topology precisely when for each $x\in X$, $f_n(x)$ converges to $f(x)$ in $Y$.

To define the remaining two topologies we need $Y$ to have a metric, which we denote by $d$. For any arbitrary metric $l$ we denote the corresponding bounded metric as $\overline{l}$, where

$$\overline{l}(x,y)=\min \{l(x,y),1\}$$

The uniform topology is identical in the context of function spaces as it is in general product spaces, where it is generated by the uniform metric

$$\overline{\rho }(x,y)=\underset{\alpha \in X}{\sup }\overline{d}(x_{\alpha },y_{\alpha })$$

In certain types of function spaces, we can alternatively obtain the uniform metric via the sup metric

$$\rho (f,g)=\underset{x\in X}{\sup }d(f(x),g(x))$$

and letting $\overline{\rho }$ be the bounded metric of the sup metric $\rho$ as the notation suggests. I say “certain types” here because for the sup metric to be well-defined we need assumptions such as $X$ being compact. A sequence $f_n$ converges to $f$ in the uniform topology precisely when $f_n$ uniformly converges to $f$. Consequently, from familiar results in analysis we can see that $\mathcal{C}(X,Y)$ is closed in $X^Y$ with the uniform topology (also recall that $\mathcal{C}(X,Y)$ is not closed in the pointwise convergence topology). Let $C$ be a compact subset of $X$. Given $f\in X^Y$ and $ϵ>0$, let

$$B_C(f,ϵ)=\{g\in X^Y|\underset{x\in C}{\sup }d(f(x),g(x))<ϵ\}$$

The sets $B_C(f,ϵ)$ form a basis over $X^Y$ that generates the topology which we call the topology of compact convergence. We make the general observation that $f_n$ converges to $f$ under the compact convergence topology precisely when the restrictions $f_n{|}_C\to f{|}_C$ uniformly for all compact subsets $C\subset X$.

Although the pointwise convergence and uniform topologies are quite familiar and thus have straightforward properties, the compact-open and compact convergence topologies require more cautious consideration. We prove some general properties.

Theorem 1) Let $X$ be a space and $(Y,d)$ be a metric space. For the function space $X^Y$, one has the following inclusions of topologies:

$$\text{(uniform)}\supset \text{(compact convergence)}\supset \text{(pointwise convergence)}$$

If $X$ is compact the first two coincide, and if $X$ is discrete the second two coincide.

Proof) Trivial.

The compact convergence topology is thus in between the uniform and pointwise convergence topologies in terms of fineness. It is, in fact, just the right amount of fineness for topological studies of function spaces as it preserves a lot of the clean properties of uniform convergence without being too restrictive. One of these is that the continuous function set is closed, under the mild assumption that $X$ is compactly generated. Recall $X$ is compactly generated if, whenever $A\subset X$ and $A\cap C$ is open in $C$ for all compact subsets $C$ of $X$, $A$ is open in $X$.

Theorem 2) Let $X$ be a compactly generated space and $(Y,d)$ be a metric space. Then $\mathcal{C}(X,Y)$ is closed in $X^Y$ under the topology of compact convergence.

Proof) Assume that $f\in \overline{\mathcal{C}(X,Y)}$, and let $C$ be an arbitrary compact subset of $X$. Then, for any $n$ we can choose some continuous function $f_n\in B_C(f,1/n)$, such that $f_n{|}_C$ uniformly converges to $f{|}_C$. We thus conclude that $f{|}_C$ is continuous for all compact subsets $C\subset X$. Then, for any open set $V\subset Y$, $f{|}_C^{-1}(V)=f^{-1}(V)\cap C$ is open in $C$ for any compact subset $C\subset X$, which implies $f^{-1}(V)$ is open in $X$ as $X$ is compactly generated. Thus, $f$ is continuous and we are done.

The one qualm we have about the topology of compact convergence is that it requires $Y$ to have a metric structure, whereas we want as much generality as possible. The compact-open topology is precisely the solution to this issue, as it is a metric-independent topology that coincides with the compact convergence topology as long as we are working in $\mathcal{C}(X,Y)$.

Theorem 3) Let $X$ be a space and $(Y,d)$ be a metric space. The compact-open topology and topology of compact convergence coincide on $\mathcal{C}(X,Y)$.

Proof) Note that the following discussion fully takes place in $\mathcal{C}(X,Y)$. We first prove that the compact convergence topology is finer than the compact-open topology. It suffices to show, for any compact subset $C\subset X$ and open set $U\subset Y$, $S(C,U)$ is open in the compact convergence topology. Consider any element $f\in S(C,U)$. As $f$ is continuous, $f(C)$ is compact, and thus there is some $ϵ>0$ such that the $ϵ$-neighborhood $B(f(C),ϵ)\subset U$. Thus, the basis element $B_C(f,ϵ)$ of the compact convergence topology satisfies $f\in B_C(f,ϵ)\subset S(C,U)$. We are done.

We now show the compact-open topology is finer than the compact convergence topology, which requires more precision. Let $B_C(f,ϵ)$ be an arbitrary basis element of the compact convergence topology, where $C\subset X$ is compact. For any $x\in C$, consider the ball $B(f(x),ϵ/4)$ in $Y$ and the neighborhood $U_x=f^{-1}(B(f(x),ϵ/4))$ of $x$ in $X$. As $C$ is compact, there are finitely many such neighborhoods $U_{x_1},U_{x_2},\dots ,U_{x_k}$ that cover $C$. Now consider the compact sets $V_{x_i}=\overline{U_{x_i}}\cap C$; $f(V_{x_i})\subset B(f(x),ϵ/3)$ as $f(V_{x_i})\subset \overline{f(U_{x_i})}=\overline{B(f(x_i),ϵ/4)}\subset B(f(x_i),ϵ/3)$. Thus, the set ${\bigcap }_{i=1}^{k}S(V_{x_i},B(f(x_i),ϵ/3))$ is an open set in the compact-open topology that contains $f$ and is a subset of $B_C(f,ϵ)$. We are done.

Theorem 4) Let $Y$ be a metric space. The compact convergence topology on $\mathcal{C}(X,Y)$ does not depend on the metric of $Y$. Therefore, if $X$ is compact, the uniform topology on $\mathcal{C}(X,Y)$ does not depend on the metric of $Y$.

Proof) This is an immediate corollary of Theorem 3.

We show two more theorems that illustrate the usefulness of the compact-open topology.

Theorem 5) Let $X$ be locally compact Hausdorff; let $\mathcal{C}(X,Y)$ have the compact-open topology. Then, the evaluation map $e:X\times \mathcal{C}(X,Y)\to Y$ defined as $e:(x,f)\mapsto f(x)$ is continuous.

Proof) Recall that locally compact Hausdorff spaces are regular. Let $U$ be a neighborhood of $f(x)$ in $Y$. By local compactness, there is a neighborhood $V$ and compact set $C\subset X$ such that $x\in V\subset C$; as $X$ is Hausdorff, $C$ is closed. Now take a neighborhood $W$ of $x$ such that $x\in W$ and $\overline{W}\subset V\cap f^{-1}(U)$; $\overline{W}$ is clearly compact. Thus, the neighborhood $W\times S(\overline{W},U)$ of $(x,f)$ in $X\times \mathcal{C}(X,Y)$ maps into $U$, and we are done.

Given a function $f:X\times Z\to Y$ such that $f(x,z_0)$ is continuous as a function on $X$ for any fixed $z_0\in Z$, we can define a corresponding function $F:Z\to \mathcal{C}(X,Y)$ such that $(F(z))(x)=f(x,z)$. We call this the map of $Z$ into $\mathcal{C}(X,Y)$ induced by $f$. Similarly, we can start from a map $Z\to \mathcal{C}(X,Y)$ to obtain a function $X\times Z\to Y$.

Theorem 6) Let $X$, $Y$, and $Z$ be spaces; give $\mathcal{C}(X,Y)$ the compact-open topology. If $f:X\times Z\to Y$ is continuous, then so is the induced function $F:Z\to \mathcal{C}(X,Y)$. The converse holds if $X$ is locally compact Hausdorff.

Proof) Take any $z_0\in Z$ and an arbitrary basis element $S(C,U)$ that contains $F(z_0)$. It suffices to obtain a neighborhood of $z$ that maps into $S(C,U)$. Consider the open set $V=f^{-1}(U)$ in $C\times Z$ with the subspace topology; the line $C\times \{z_0\}\subset V$. By the tube lemma, there must be a neighborhood $W$ of $z_0$ such that $C\times W\subset V$. This $W$ satisfies $F(W)\subset S(C,U)$, as desired.

For the converse, the map $f:X\times Z\to Y$ can be represented as $f=e\circ h$ where $h:X\times Z\to X\times \mathcal{C}(X,Y)$ is $h=i_X\times F$. $h$ is clearly continuous by hypothesis, and $e$ is continuous as $X$ is locally compact Hausdorff. We are done.