# Compact open topology on space of continuous maps from a discrete space is a product

I'm really confused with the statement.

let $X$ be a set endowed with the discrete topology and let $Y$ be any topological space. Show that $M(X,Y)$, the space of continuous maps from $X$ to $Y$ endowed with the compact open topology, is homeomorphic to the topological product $\prod_{x \in X} Y_x$, $Y_x = Y$.

Can you please explain what we really have to prove?

• Topological product of what? $X$ and $Y$? – Brian Fitzpatrick Mar 6 '14 at 5:57
• First, $M(X,Y)=Y^X$ as sets (why?). Then write down the subbasic open sets for both topologies (what are the compact subsets of $X$?), see that they are the same, thus define the same topology. – Olivier Bégassat Mar 6 '14 at 6:16
• M(X,Y) is subset of $Y^X$ that consists of all continuos maps – Knight Mar 6 '14 at 7:00

## 1 Answer

First, $M(X,Y)$ equals $Y^X=\prod_{x\in X} Y$, the set of all functions from $X$ to $Y$ since ever function on a discrete space is continuous.
Now the subbasis for the product topology on $Y^X$ consists of all $$e_x^{-1}(U)=\{(f(x))_{x\in X}\mid e_x(f)=f(x)\in U\}$$ ranging over the open subsets $U$ of $X$ and the elements $x$ of $X$. The subbasis for the compact-open topology on $M(X,Y)$ consists of the sets $$(K,U)=\{f:X\to Y\mid f(x)\in U\forall x\in K\}$$ for all compact $K$ and open $U$.
Can you show that each $e^{-1}_x(U)$ has the form $(K,U)$ for some compact $K$, and conversely, that each $(K,U)$ is an intersection of $e_x^{-1}(U)$ over all $x$ in a finite set $\{x_1,...,x_n\}$ ?