4
$\begingroup$

Munkres - Topology p. 283

Definition

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..

$\endgroup$
  • 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 '13 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. $\endgroup$ – Moishe Kohan Nov 10 '13 at 10:26
4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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)? $\endgroup$ – Jonathan Y. Nov 10 '13 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. $\endgroup$ – Brian M. Scott Nov 10 '13 at 10:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.