Let $(X, \mathscr{T})$ be a topological space and $(Y,d)$ be a metric space.
Recall that the compact-open topology $\mathscr{T}_{co}$ on $Y^X$ is generated by the subbase
$$ \mathscr{S} = \{ S(C, V) \mid C \text{ compact of } X, V \in \mathscr{T}_d\} $$
where $S(A,B) = \{f : X \to Y \mid f(A) \subseteq B\}$ for $A \in \mathscr{T}$, $B \in \mathscr{T}_d$
I wish to prove or disprove that a limit of bounded functions for $X = I$, $Y = \mathbb{R}$ is bounded. My counterexample of choice would be $f_n(x) = \min\{n, f(x)\}$ for some unbounded $f$ but I don't think it works here. Could anyone help me solve this exercise? Thanks in advance.
Note: I know that the compact convergence topology coincides with the compact-open topology on $\mathscr{C}(X,Y)$, but I don't see a reason why this should happen for bounded functions. Therefore, any advice regarding the same problem, but for the compact convergence topology is also appreciated.
If $X$ is compact then the statement is true (the uniform limit of a sequence of bounded functions is bounded).
If $X$ is not compact then it might fail. Consider some compact exhaustion $\mathbb{R}=\bigcup \limits _{n \geq 0} K_n$. Consider $f_n = x, x \in K_n$ and $f_n=0, x \notin K_n$. Clearly, $f_n(x) \to x \space \forall n$ (pointwise convergence), so the limit is unbounded. Let us show that the convergence happens not only pointwisely, but in the compact-open topology.
Let $K$ be some compact. Since the $K_n$-s exhaust $\mathbb{R}$, there is an $n_0$ such that for $n \geq n_0$ $K \subset K_n$. Then $\sup \limits _{x \in K} |f_n (x) - x| = \sup \limits _{x \in K_n} |f_n (x) - x| = \sup \limits _{x \in K_n} |x - x| = 0$, so $f_n$ converges uniformly to the function $x$ on any arbitrary $K$.
Therefore, for a specific non-compact space ($\mathbb{R}$) we have produced a sequence of bounded functions that tends to an unbounded one in the compact-open topology.
In conclusion, boundedness is not preserved in the compact-open convergence on non-compact spaces. It is on compact spaces.