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 $f(\infty)=0$).



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> Is $C\_{0}(X)$ always Frechet-Urysohn with respect to the compact-open topology (when $X$ is locally compact)?
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Is $C_0(X) $ Frechet-Urysohn with respect to the compact-open topology?
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 $f = 1\_\alpha \in A$.


