Is $C_0(X) $ Frechet-Urysohn with respect to the compact-open topology? Recall that a topological space is called Frechet-Urysohn if the operations of closure and sequential closure coincide.
Let $X$ be a locally compact Hausdorff space. It is known that $C(X)$ is not Frechet-Urysohn with respect to the compact-open topology. Indeed, it was proven in the paper Hernández, Mazón - On the sequential spaces of continuous functions, which I cannot access, that if $X$ is first countable, then $C(X)$ is  Frechet-Urysohn iff $X$ is hemi-compact (which means that there is a sequence of compact sets such that every compact is contain in an element of the sequence). My question however is not about the whole  $C(X)$, but its subset - $C_{0}(X)$.
Recall that $C_{0}(X)$ is the set of all $f\in C(X)$ that "vanish at infinity", i.e. for every $\varepsilon>0$ there is a compact $K\subset X$ such that $|f(x)|<\varepsilon$ as soon as $x\not\in K$ (equivalently, $f$ is continuously extended to the one-point compactification of $X$ by setting $f(\infty)=0$).

Is $C_{0}(X)$ always Frechet-Urysohn with respect to the compact-open topology (when $X$ is locally compact)?

 A: In general, $C_0(X)$ need not be Frechet-Urysohn.
For a counterexample, let $X = \omega_1$ be the first uncountable ordinal with its order topology.  Let $S \subset \omega_1$ be the set of successor ordinals in $\omega_1$.  For $\alpha \in S$, let $1_\alpha : \omega_1 \to \mathbb{R}$ be the function which is $1$ at $\alpha$ and $0$ elsewhere; then each $1_\alpha$ is continuous and compactly supported.  Consider the set $A = \{1_\alpha : \alpha \in S\} \subset C_0(\omega_1)$.  I claim $0$ is in the closure of $A$ but not in the sequential closure.
Note that a basis for the open neighborhoods of $0$ in the compact-open topology is given by the sets $U_{\beta, \epsilon} = \{ f : |f(x)| < \epsilon, \forall x \le \beta\}$, where $\beta \in \omega_1$ and $\epsilon > 0$.  (The intervals $[0, \beta]$ are compact, and every compact set is contained in such an interval.)  Then $1_{\beta+1} \in U_{\beta, \epsilon}$, so that every open neighborhood of $0$ contains a point of $A$, and thus $0$ is in the closure of $A$.
On the other hand, suppose $1_{\alpha_n}$ is a sequence in $A$.  Let $\beta = \sup_n \alpha_n \in \omega_1$.  Then none of the $1_{\alpha_n}$ are in $U_{\beta, 1/2}$, so $1_{\alpha_n}$ does not converge to $0$, and so $0$ is not in the sequential closure of $A$.
In fact, $A$ is sequentially closed.  Suppose $1_{\alpha_n} \in A$ converges to some $f \in C_0(\omega_1)$.  Since $\omega_1$ is well ordered, we can pass to a subsequence so that $\alpha_n$ is nondecreasing.  If it does not stabilize, then $1_{\alpha_n} \to 0$ pointwise, which contradicts uniform convergence on the compact set $[0, \sup_n \alpha_n]$.  So $\alpha_n$ must stabilize at some $\alpha$ which means $f = 1_\alpha \in A$.
