# Are all subspaces of the rationals order topologies (with respect to some linear order)? All closed subspaces?

This is a follow-up to this question. Let $Y \subset \mathbb{Q}$, and give $Y$ the subspace topology. Is there necessarily a linear ordering $<$ of $Y$ (possibly different from the order inherited from $\mathbb{Q}$) which induces the topology on $Y$? What if $Y$ is assumed to be closed in $\mathbb{Q}$?

Some discussion: let $L$ be the set of limit points of $Y$. Let $I$ be the set of isolated points of $Y$. Note that $I$ must have some limit points (in $L$). Otherwise, $L$ is perfect so $L \cong \mathbb{Q} \cong \mathbb{Q} \cap [0,1]$ by a theorem of Sierpiński. Also we have $Y$ homeomorphic to the disjoint union of $L$ and $I$ in this case. It is easy to see that the disjoint union of $\mathbb{Q} \cap [0,1]$ with a discrete space can be realized as an order topology. So, $I$ needs to have some limit points, eg. something like $Y = \{0,1,1/2,1/3,1/4,\ldots\}$ (but the latter is an order topology).

-

Yes, there is. Proposition 3.2 of Bennett & Lutzer, A shorter proof of a theorem on hereditarily orderable spaces, says (among other things) that a metrizable space is hereditarily orderable iff it is orderable and totally disconnected. $\mathbb{Q}$ is certainly metrizable, orderable, and totally disconnected, so it must be hereditarily orderable.
@Asaf: Yes, for example the subspace $\{0\}\cup(1,2)$ has a topology which can't be induced by any linear order. – LostInMath Nov 19 '11 at 13:38