# Prove that the space of continuous functions $C(X,Y)$ with the topology of compact convergence is Hausdorff.

Let $$X$$ be a topological space and $$Y$$ be a metric space. Prove that the space of continuous functions $$C(X, Y )$$ with the topology of compact convergence is Hausdorff.

The compact convergence topology is generated by the collection of sets $$B_C(f,ϵ)=\{g∈Y^X: \text{sup }_{x∈C} d(f(x),g(x))<ϵ\}$$, with compact $$C⊂X$$.

I know to show a topology is Hausdorff we need to show that for any two distinct elements in the topology we can find neighborhoods of those elements that do not intersect, but I am not sure how to do this for the compact convergence topology. The functions and supremum are throwing me off.

Suppose $$f \neq g$$. We know that there must be some $$p \in X$$ with $$f(p) \neq g(p)$$. Define $$\varepsilon = \frac{d(f(p),g(p))}{2} >0$$, where $$d$$ is the metric on $$Y$$.

Now define $$C=\{p\}$$, which is finite, so compact. A supremum of $$d(f(x), g(x))$$ with $$x \in C$$ is then just the single value $$d(f(p), g(p))$$.

Then $$f \in B_C(f,\varepsilon) = \{h \in Y^X: d(f(p), h(p)) < \varepsilon\}$$ and $$g \in B_C(g,\varepsilon) = \{h \in Y^X: d(g(p), h(p)) < \varepsilon\}$$.

If $$h \in B_C(f, \varepsilon) \cap B(g, \varepsilon)$$ existed, then both

$$d(h(p), f(p)) < \varepsilon \text{, and } d(h(p), g(p)) < \varepsilon$$

and so $$d(f(p), g(p)) \le d(f(p), h(p)) + d(h(p), g(p)) < 2\varepsilon = d(f(p), g(p)$$

which is a contradiction. So $$B_C(f, \varepsilon) \cap B(g, \varepsilon) = \emptyset$$ and $$C(X,Y)$$ is Hausdorff.

Hint: Singleton sets $$\{x\}$$ are always compact