7.3Pointwise and Compact Convergence

Updated 31 Aug 2026Chapter (PDF)

[7.3.1]Definition(Topology of Pointwise Convergence)#

Let $X$ be a topological space. Given any $x \in X$ and open set $U$ of the space $Y$ , let

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

The collection of all $S(x, U)$ forms a subbasis for a topology on $Y^X$ called the topology of pointwise convergence.

Let's consider the basis elements for this topology, which are simply finite intersections of subbasis elements. For some $f: X \to Y$ , a basis element about $f$ would comprise functions $g$ that are “close” to $f$ at finitely many points. For example, let

B = \bigcap_{k = 1}^n S(x_k, U_k)

be an arbitrary basis element about an arbitrary function $f \in Y^X$ . For $g \in B$ to be true, we need that $g(x_k) \in U_k$ for each $k \in \{1, \dots, n\}$ . Of course, $f \in B$ as well, so this is the notion for which we say the functions are “close” to eachother.

Note this topology on $Y^X$ is the exact same as the product topology. If we replace $X$ with an indexing set $\Lambda$ and denote a general element of $\Lambda$ by $\lambda$ , then the set $S(\lambda, U)$ simply comprises the functions $\vb{x}: \Lambda \to Y$ such that $\vb{x}(\lambda) = x_\lambda \in U$ . This is precisely $\pi_\lambda^{-1}(U)$ of $Y^\Lambda$ , and so subbasis elements agree with the product topology.

[7.3.2]Theorem#

Let $Y^X$ be a topological space with the topology of pointwise convergence (or otherwise the usual product topology). Then a sequence of functions $f_n$ converges to a function $f$ if and only if, for every $x \in X$ , we have that $f_n(x) \to f(x)$ . That is, $f_n \to f$ if and only if $f_n$ converges pointwise to $f$ .

[7.3.3]Definition(Topology of Compact Convergence)#

Let $X$ be a topological space and $(Y, d)$ a metric space. Given some $f \in Y^X$ , a compact subset $C \subseteq X$ , and some $\epsilon > 0$ , we define

B_C(f, \epsilon) = \Bigg\{ g: \sup_{x \in C} \{d(f(x), g(x))\} < \epsilon \Bigg\}.

The collection of all $B_C(f, \epsilon)$ forms a basis for a topology on $Y^X$ called the topology of compact convergence.

The proof that this forms a basis for a topology is immediate. The idea behind this topology is that a basis element about $f$ admits functions $g$ that are “close” to $f$ on an entire compact set, instead of finitely many points.

[7.3.4]Theorem#

Let $Y^X$ be a topological space with the topology of compact convergence. Then a sequence of functions $f_n$ converges to $f$ if and only if $f_n|_C \to f|_C$ uniformly on every compact $C \subseteq X$ .

Proof.

Recall that we say $f_n \to f$ uniformly on some $A \subseteq X$ if, for any $\epsilon > 0$ , there exists some $N \in \bZ_+$ such that $d(f_n(x), f(x)) < \epsilon$ for every $x \in A$ . In other words, the choice of $N \in \bZ_+$ must work for all points in $A$ simultaneously, as opposed to at a single point like with uniform convergence. The notion of convergence induced by this topology is that, after fixing some compact set $C$ , $f_n|_C \to f|_C$ if for any basis element $U = B_C(f, \epsilon)$ (this step involves simply choosing any $\epsilon > 0$ ), there is some $N \in \bZ_+$ such that $f_n \in U$ for every $n \ge N$ . If $f_n \in U$ , then we have that $\sup_{x \in C} d(f(x), f_n(x)) < \epsilon$ . Thus $f_n|_C \to f|_C$ is precisely what it means for $f_n|_C \to f|_C$ uniformly.

[7.3.5]Theorem(Topologies on Function Space)#

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

(\text{uniform}) \supseteq (\text{compact convergence}) \supseteq (\text{pointwise convergence}).

If $X$ is compact, the first two topologies coincide. If $X$ is discrete, the last two topologies coincide.