This is almost certainly a duplicate, but I keep seeing this result on metric spaces, not topological ones.
Let $(X,\tau)$ be a topology. A set $A\subset X$ is dense if $A\cap B\neq\emptyset$ for all $B\in\tau$. We say $(X,\tau)$ is separable if there exists a countable, dense $A\subseteq X$.
Given some $A\subseteq X$, a point $p\in A$ is an isolated point in $A$ if there exists $O\in\tau$ such that $p\in O$ and $O\cap A=\{p\}$.
I am wondering: If $X$ is separable and $A\subseteq X$, then must the set of isolated points in $A$ be at most countable? Perhaps if we add the condition that it is Hausdorff it is true.
My attempt: If $A$ has $0$ or $1$ isolated points, we are done. Otherwise, let $p_{1},p_{2}\in A$ be isolated points of $A$. Then there exist $O_{1},O_{2}\in\tau$ such that $O_{1}\cap A=\{p_{1}\}$ and $O_{2}\cap A=\{p_{2}\}$. Furthermore, because $(X,\tau)$ is Hausdorff, there exist $T_{1},T_{2}\in\tau$ such that $p_{1}\in T_{1},p_{2}\in T_{2}$, and $T_{1}\cap T_{2}=\emptyset$.
Now, because open sets are closed under finite intersection, we have that $O_{1}\cap T_{1}$ and $O_{2}\cap T_{2}$ are open, disjoint sets which have intersection $\{p_{1}\}$ and $\{p_{2}\}$ with $A$, respectively.
My idea from here is to well order some countable dense subset and use the well-ordering to choose one element from each open set around each isolated point (without using choice because we can just choose the least element). But I have yet to show that there exists a collection of disjoint open sets, one for each isolated point. I am not sure how to continue.
For example, the result is true in the reals for closed sets by Cantor-Bendixon (I think). However the proof I saw was nothing like this and the fact that I've not seen a more general statement for any set of reals seems like an indicator that it is not true. Is it true if I add more restrictions? Maybe a stronger separation axiom?
Best Answer
No, there are many counterexamples. One of my favourites is Mrówka's $\Psi$ space, which I talked about in this answer, see also this blog post for much more info.
It's basically a countable open subset $D$ of isolated points that is dense in $X$ while $X\setminus D$ is uncountable and discrete as a subspace (so all its points are isolated within that set). The rational sequence topology on $\Bbb R$ is another instance of the same idea and also works as a more elementary counterexample.
It's indeed true for metric spaces in general. If a metric space is separable, it's second countable and thus hereditarily separable and hereditary Lindelöf and both of those last properties imply that all discrete (in themselves) subspaces are at most countable, which is what you were trying to show.
A space $X$ where a discrete subspace is at most countable is said to have countable spread, denoted by $s(X) = \aleph_0$. (Separable is countable density, $d(X)=\aleph_0$, second countable is called countable weight, $w(X)=\aleph_0$, and many other so-called cardinal invariants of spaces have been defined and studied, as well as their relationships. In these terms I have given counterexamples to the hypothesis $s(X) \le d(X)$ while in metric spaces $d(X)=hd(X)$ so there $s(X) \le d(X)$ does hold.)