TITLE: Scattered hereditarily separable does not imply countable in ZFC
QUESTION [7 upvotes]: 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 $\mathsf{ZFC}$.

REPLY [8 votes]: 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.