Munkres - Topology p. 283


Let $(Y,d)$ be a metric space and $X$ be a topological space. Define $B_C(f,\epsilon)$ as the set $\{g\in Y^X : \sup\limits_{x\in C} \operatorname{d}(f(x),g(x)) < \epsilon \}$ for a given compact subspace $C$ and $\epsilon >0$ and $f\in Y^X$.

Then, the topology generated by all $B_C(f,\epsilon)$ is called the "Topology of compact convergence".

How does this is a well-defined definition?

Munkres mentioned in his book that we need some topology on $Y^X$ making $C(X,Y)$ closed which is stronger than the product topology. Then, he defined 'the topology of compact convergence' as given above.

Since he considers a topology on $Y^X$, he didn't assume functions to be continuous, hence not bounded.

Well, if functions are not continuous, then compactness of $C$ no more gurantees that $\sup_{x\in C} d(f(x),g(x))$ exists even when $C$ is nonempty, and of course it does not exist when it is empty.

Is he taking the supremum over the extended real?

Or, should i take $d$ as a bounded metric?

What would be the definition of this that makes sense?

Off the topic, i feel like munkres define topologies that nobody uses but really useful. An example is the uniform metric. And i think 'topology of compact convergence' would be the one too. There's no definition for this topology in wikipedia..

  • 2
    $\begingroup$ I think you have omitted part of the definition of the set $B_C (f,\epsilon)$. (Probably »$\sup < \epsilon$«.) $\endgroup$
    – user642796
    Nov 10, 2013 at 10:26
  • $\begingroup$ This topology is usually called topology of "uniform convergence on compacts" and is used, to be best of my knowledge, a lot, but only in the context of continuous functions. $\endgroup$ Nov 10, 2013 at 10:26

1 Answer 1


The definition given by Munkres is correct. The set $B_C(f,\epsilon)$ contains the functions $g:X\to Y$ for which $\sup_{x\in C}d\big(f(x),g(x)\big)$ exists and is less than $\epsilon$. If $g:X\to Y$ is such that the supremum doesn’t exist (or if you prefer, is infinite), then $g\notin B_C(f,\epsilon)$, that’s all.

The topology of compact covergence is defined in Wikipedia; the definition is given in terms of which sequences of functions converge rather than directly in terms of the topology, but if you compare it with Theorem $46.2$ in Munkres, you’ll see that it’s the same topology.

Both the uniform topology and the topology of compact convergence are extremely useful and widely used.

  • $\begingroup$ BrianM.Scott, if you don't mind elaborating (and assuming the reference can be explained in the available space), does compact convergence truly induce a topology, and if so, why is it determined by convergent sequences (is it, as the notation suggests, metrizable)? $\endgroup$ Nov 10, 2013 at 10:36
  • 1
    $\begingroup$ @Jonathan: It’s not hard to check that the family of sets $B_C(f,\epsilon)$, where $f$ ranges over all functions from $X$ to $Y$, $C$ over all compact subsets of $X$, and $\epsilon$ over all positive reals, is a base for a topology on ${}^XY$. The topology is not in general metrizable, but in most settings of interest it is determined by the convergent sequences, though I don’t offhand remember exactly what conditions are required to ensure that. $\endgroup$ Nov 10, 2013 at 10:55
  • $\begingroup$ @BrianM.Scott 'In most setting, topology determined by convergent sequences'. Will it occur in case of topology of compact convergence? I am unable to see it $\endgroup$
    – Sushil
    May 12, 2021 at 15:50
  • $\begingroup$ @Sushil: In that comment I was talking about the topology of compact convergence, so the answer is sometimes. It’s been years since I worked with these ideas, so I can’t say any more than I did back then, but you can get more information from Daniel Fischer’s answer to this question. $\endgroup$ May 12, 2021 at 21:09
  • $\begingroup$ @BrianM.Scott Thanks a lot sir. I misunderstood your comment that sometimes topology on general set can be described by sequence convergence and I though topology of compact convergence of this type. Thanks for clarification and answer of Danial Fisher $\endgroup$
    – Sushil
    May 13, 2021 at 17:18

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