# When is an ordered space scattered?

There is a concept of scattered in both order theory and topology.

• A topological space $X$ is scattered if every nonempty subspace has an isolated point.
• A linearly ordered set $( X , < )$ is scattered if it has no densely ordered subsets of size at least $2$ (that is, for each $A \subseteq X$ containing at least two elements there are $a < b$ in $A$ such that there is no $x \in A$ with $a < x < b$).

Recall that given a linearly ordered set $( X , < )$, the order topology on $X$ induced by $<$ is generated by the subbasis consisting of all set of the form \begin{align} ( \leftarrow , a ) &:= \{ x \in X : x < a \} \\ ( a , \rightarrow ) &:= \{ x \in X : x > a \} \end{align} for $a \in X$.

Now, given a scattered linear order $(X , < )$, the order topology on $X$ is scattered, however the converse does not hold. Wikipedia gives the example of the lexicographic order on $\mathbb{Q} \times \mathbb{Z}$.

• Clearly $\mathbb{Q} \times \{ 0 \}$ is a densely ordered subset, so it is not a scattered order.
• However the order topology on $\mathbb{Q} \times \mathbb{Z}$ is discrete: for each $(q,n) \in \mathbb{Q} \times \mathbb{Z}$ we have that $(\,(q,n-1),(q,n+1)\,) = \{ (q,n) \}$ is open.

Question. Is there a "nice" order-theoretic characterisation of when the order topology of a linearly ordered set is scattered?

• I suspect not, but papers by Steve Purisch would be one place to look. Jun 6, 2015 at 18:59