Compact-Open topology for the space of all continuous paths

By | October 25, 2019

As i mentioned in Topological Complexity, M.Farber shows the space of all continuous paths in space $X$ by $PX$ and equip the path space  $PX$  with compact-open topology. But why?

In this note i try to account compact-open topology and another one called compact convergence topology and obviously their relations.

Compact-Open topology

When we have a space of maps, more precisely in this case, a space of continuous functions (paths), and we want to exert a topology on this space, it is natural to considering compact-open topology. But what is it exactly?

Here i rewrite the definitiion of compact-open topology from “topology, A first course by Munkres”:

Let $X$ and $Y$ be topological spaces. If $C$ is a compact subset of $X$ and $U$ is an open subset of $Y$, define

$S(C,U) =\{f \mid f \in \mathcal C(X,Y) \thinspace and \thinspace f(C)\subset U \}$

The sets $S(C,U)$ form a subbasis for a topology on $\mathcal C(X,Y)$ called the compact-open topology.

Compact convergence topology

Again from “topology, A first course by Munkres” we have:

Let $(Y,d)$ be a metric space; let $X$ be a topological space. Given an element $f$ of $Y^X$, a compact subset $C$ in $X$, and a number $\epsilon > 0$, let $B_{c}(f,e)$ denote the set of all those elements $g$ of $Y^X$ for which

$lub\{d(f(x),g(x)) \mid x \in C \} < \epsilon$

The sets $B_{c}(f,e)$ form a basis for a topology on $Y^X$. It is called the topology of compact convergence (or sometimes the “topology of uniform convergence on compact sets”).

Coincidence of Compact convergence and Compact-Open topology

Theorem 5.1 in the book noted above is as below:

Let $X$ be a space and let $(Y,d)$ be a metric space. For the space $\mathcal C(X,Y)$, the compact-open topology and topology of compact convergence coincide.

With this theorem we can understand the continuity of functions showed by $s_{i}$ in the notion of Topological Complexity, better and better. In fact with compact convergence topology on $X^I$, we can conceive closing of two paths more easier, because we can assume a metric on $X$.

But on the other hand we know that this metric is not important, means the structured topology doesn’t depend on the metric of $X$.