Compact-Open Topology and Discrete Spaces
it is true in general that $C(X,Y)$, the set of all continuous maps from $X$ to $Y$ equipped with the compact-open topology, is discrete given that $Y$ is discrete?
It certainly holds when $Y$ is just a singleton.
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$\begingroup$First, just to fix some notation, given topological spaces $X,Y$, a compact $K \subseteq X$ and an open $V \subseteq Y$, by $U[K,V]$ I will denote the set $\{ f \in C(X,Y): f[K] \subseteq V \}$. Recall that the family of all these $U[K,V]$ is a subbasis for the cimpact-open topology on $C(X,Y)$. (Meaning that the family of all finite intersections of these $U[K,V]$ sets is a basis for the compact-open topology.)
Let $X = Y = \mathbb{N}$ with the discrete topology. I claim that $C(X,Y)$ is not discrete. Note that the compact subsets of $X$ are just the finite sets. Given finitely many compact (finite) $K_1 , \ldots , K_n \subseteq X$ and $V_1 , \ldots , V_n \subseteq Y$ and consider $$U[K_1,V_1] \cap \cdots \cap U[K_n,V_n] = \{ f \in C(X,Y) : f[K_i] \subseteq V_i \text{ for each }i \leq n \}.$$ If this set is nonempty (e.g., if the $K_i$ are pairwise disjoint), pick any $f$ in it. Note that $K_1 \cup \cdots \cup K_n$ is finite, and so $X \setminus ( K_1 \cup \cdots \cup K_n ) \neq \varnothing$, Define a function $g : X \to Y$ by $$g(x) = \begin{cases} f(x), &\text{if }x \in K_1 \cup \cdots \cup K_n \\ f(x)+1, &\text{otherwise.} \end{cases}$$ It is easy to see that $g \in C(X,Y)$, $g \neq f$ and $g \in U[K_1,V_1] \cap \cdots \cap U[K_n,V_n]$. Therefore no finite intersection of the $U[K,V]$ sets can be a singleton, and since these sets are a basis for the compact-open topology, no singleton is open in $C(X,Y)$, meaning that $C(X,Y)$ is not discrete.
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