1
$\begingroup$

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)?

$\endgroup$
2
  • 1
    $\begingroup$ 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. $\endgroup$ – Nate Eldredge Jan 16 at 21:47
  • $\begingroup$ @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. $\endgroup$ – erz Jan 16 at 22:32
2
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.