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?


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.)

  • $\begingroup$ The first one I thought of was a Souslin (Suslin) line but it's outside of ZFC. Incidentally, the last problem on linear spaces in Engelking's General Topology (Prove that a linear space is hereditarily collection-wise normal) makes proving some much earlier problems much easier: That for a linear space $X$ we have $hd(X)=d(X)$ and $hc(X)=c(X)=hl(X)$. $\endgroup$ – DanielWainfleet Aug 25 at 5:53
  • $\begingroup$ @DanielWainfleet A Suslin line does obey $d(X)=hd(X)$ because it's an ordered space, so within ordered spaces there are no counterexamples. Your "linear space" also means LOTS (linearly ordered topological space) I suppose. $d(X)=hd(X)$ holds for monotonically normal spaces more generally. $\endgroup$ – Henno Brandsma Aug 25 at 5:57
  • $\begingroup$ @DanielWainfleet linear space is a TVS for me (and most other topologists). $\endgroup$ – Henno Brandsma Aug 25 at 6:05
  • $\begingroup$ Yes I should have said LOTS. $\endgroup$ – DanielWainfleet Aug 26 at 23:49

A simple example is the Sorgenfrey plane, i.e., the plane $\mathbb R\times\mathbb R$ with the topology generated by the half-open rectangles $[a,b)\times[c,d)$. The set $\mathbb Q\times\mathbb Q$ of all rational points is a countable dense set, and the anti-diagonal $\{(x,-x):x\in\mathbb R\}$ is an uncountable discrete closed subset.

Another nice example is the compact Hausdorff space $\{0,1\}^\mathfrak c$, the product of continuum many two-point discrete spaces, which can be shown to be separable. The set of all points with a single nonzero coordinate is an uncountable discrete subset.


The Niemytzki plane (or Moore plane) is a counterexample.

  • The family of all pairs of rational numbers is dense in the space, so it is separable.
  • The $x$-axis $\{ (x,0) : x \in \mathbb{R} \}$ is an uncountable discrete subspace.

The only property I can think of at the moment which would imply that every subspace has countably many isolated points is hereditary separability, meaning that all subspaces are separable. (If $X$ has a subset $A$ with uncountably many isolated points, then $B \subseteq A \subseteq X$ consisting of the isolated points of $A$ would be an uncountable discrete subspace of $X$, which cannot be separable.) I am not certain at the moment if this is equivalent to having all subsets having countably many isolated points, but I would doubt it.


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.