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?

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

1 Answer 1


I don't know about the general case, but there is some work on this from several years ago. In "When are order scattered and topologically scattered the same?" (Annals of Discrete Mathematics 23 (1984), pp 61-80), it was shown that for distributive algebraic lattices, which can be endowed with a compact Hausdorff topology called the Lawson topology, order scattered and topologically scattered are equivalent. In particular, if the lattice is a total order, then they are equivalent. It was also noted that topologically scattered compact pospaces are order scattered, but the converse fails for some nondistributive algebraic lattices.


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