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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.

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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.

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

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