# 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)?

• What about taking $X = \omega_1$, the first uncountable ordinal with its order topology? If $A \subset C_0(\omega_1)$ is the set of all $1_{\{\alpha\}}$ where $\alpha$ is a successor, then I think the constant function 0 is in the closure of $A$ but not in the sequential closure. – Nate Eldredge Jan 16 at 21:47
• @NateEldredge it seem to work, thank you! If you post this as an answer (perhaps with some details) I'd be glad to accept it. – erz Jan 16 at 22:32

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 $$f = 1_\alpha \in A$$.