# Example of topological space where there is a point and a subset $A$: $x \in \overline A$, but no sequence in $A$ converging to $x$?

It is known that if a space $X$ is metricable then for any subset $A$ of $X$ and a point $x \in \overline A$ there is a sequence of points in $A$ converging to $x$.

I wonder if there is an example of a topological space $X$ (not metricable then) where this doesn't always hold?

• Are you requiring $x\in \overline{A}\setminus A$ (not stated as such)? Are you allowing the sequence to consist of $x_n=\{x,x,x,\dots\}$? – JohnD Nov 25 '14 at 17:12
• @JohnD: wouldn't you allow these? In a discrete space, it is necessary to have those in order to have equivalence between $\overline A$ and limits of sequences. It would be silly to discard this type of sequence, and withdraw the generality of the statement "In any metrizable space X, ...". – zarathustra Nov 25 '14 at 17:18
There are many such spaces. One simple Hausdorff example is the Arens-Fort space, which is fairly clearly described in the linked article. In that space let $A$ be everything except the point $\langle 0,0\rangle$; then $\langle 0,0\rangle\in\operatorname{cl}A$, but no sequence in $A$ converges to $X$.
An even simpler example, but one that isn’t Hausdorff, is an uncountable set $X$ with the co-countable topology: the open sets are $\varnothing$ and the complements of countable sets. Let $p\in X$ be arbitrary, and let $A=X\setminus\{p\}$; then $p\in\operatorname{cl}A$, but no sequence in $A$ converges to $p$. (In fact, the only convergent sequences in $X$ are the trivial ones, i.e., the ones that are constant from some term on.)
There are also nice spaces with this property, e.g., compact Hausdorff spaces. If we let $X$ be the set of all ordinal numbers less than or equal to $\omega_1$, the first uncountable ordinal, and give $X$ the order topology, then $X$ is a compact Hausdorff space, $\omega_1\in\operatorname{cl}[0,\omega_1)$, and no sequence in $[0,\omega_1)$ converges to $\omega_1$.