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Recall that a topological space $X$ is scattered if and only if every non-empty subset $Y$ of $X$ contains at least one point which is isolated in $Y$. Consider the statement: "Every scattered hereditarily separable space is countable".

In this book it is mentioned that, under $\mathsf{CH}$, K. Kunen constructed a compact, scattered, hereditarily separable space of size $\aleph_1$. My question is if a "real" example of a scattered, hereditarily separable, uncountable space is known, or if this statement is independent of $\mathsf{ZFC}$.

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  • $\begingroup$ Wouldn't such a space give a ZFC space which is hereditarily separable but not Lindelof? (Since each scattered level is countable, there are uncountably many scattered levels; take a subspace of scattered height $\omega_1$ and whose top level is empty. The cover $\{U_\alpha : \alpha < \omega_1\}$, where $U_\alpha$ is the union of the levels below the $\alpha$th level, has no countable subcover.) It is known to be consistent that no such space exists. $\endgroup$
    – Anonymous
    Commented Jan 18, 2022 at 16:01
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    $\begingroup$ Your argument is right: every scattered, hereditarily separable, uncountable space is not Lindelöf. On the other hand, in your comment about consistency, I think you are referring to $S$-spaces ($T_3$, hereditarily separable non-Lindelöf). As I recall, Stevo Todorčević showed that under PFA there are no $S$-spaces. However, there are examples of hereditarily separable, non-Lindelöf, non-regular Hausdorff spaces in ZFC. Since we didn't require any separation axioms, I think you answer doesn't settle the question. $\endgroup$
    – Peluso
    Commented Jan 18, 2022 at 21:05
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    $\begingroup$ I apologize--I didn't realize that you did not want to make assumptions about separation axioms. In that case, I think you can get an example by letting $X$ be an uncountable ordinal and giving it the topology having for a base sets of the form $[0,\alpha)$. Then if $S$ is any non-empty subset of $X$ and $p$ is the smallest element of $S$, the set $\{p\}$ is dense in $S$ and the point $p$ is isolated in $S$. $\endgroup$
    – Anonymous
    Commented Jan 19, 2022 at 0:50
  • $\begingroup$ Nice! I kind of convinced myself that it was a difficult thing to construct after reading about Kunen's thingy. If you want to, you can write a brief answer so that this question does not remain unanswered. Just out of curiosity and since the topic came up, do you know any $T_2$ (or $T_1$) examples? $\endgroup$
    – Peluso
    Commented Jan 19, 2022 at 2:55
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    $\begingroup$ I haven't thought about $T_2$ or $T_1$ examples, but you might be able to get a $T_1$ example by using almost the same example as before, but taking a base to be sets of the form $[0,\alpha) \setminus F$ where $F$ is finite. $\endgroup$
    – Anonymous
    Commented Jan 19, 2022 at 13:54

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As clarified in the comments, the existence of a regular S-space is independent of ZFC, but a "real" $T_2$ example can be constructed by taking a well-ordering of a set of reals in order type $\omega_1$ and refining the Euclidean topology by declaring initial segments open. This example is $T_2$ since it refines a $T_2$ topology and is still hereditarily separable (any subset has an initial segment that is dense in the Euclidean topology and this initial segment is still dense in the stronger topology). It is not Lindelof since the cover by initial segments has no countable subcover, and it is scattered since any subset has a minimal element which is isolated in that subspace. This example is described in Mary Ellen Rudin's excellent book "Lectures on Set Theoretic Topology" published by the AMS.

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    $\begingroup$ Hi Paul -- welcome to MathOverflow! $\endgroup$
    – Will Brian
    Commented Jan 20, 2022 at 12:19

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