1
$\begingroup$

I'm trying to show that there is an uncountable ordered set such that all of its proper initial segments are similar to $\mathbb Q$ or to $\mathbb Q \cap (0,1]$.

I can show that any countable densely ordered set is similar to one of the sets $\mathbb Q \cap (0,1)$, $\mathbb Q \cap (0,1]$, $\mathbb Q \cap [0,1)$, or $\mathbb Q \cap [0,1]$ (depending if it has a first or last element), but not sure what to do from there.

$\endgroup$
2
$\begingroup$

Hint: Let $\omega_1$ be the least uncountable ordinal and let $X = \omega_1 \times \mathbb Q$. Let $\prec$ be the strict lexicographical order on $X$, i.e. for all $(a,b), (c,d) \in X$ we let $(a,b) \prec (c,d)$ iff $a \in c$ or [$a=c$ and $b < d$]. Furthermore, let $\preceq$ be defined by $(a,b) \preceq (c,d)$ iff $(a,b) = (c,d)$ or $(a,b) \prec (c,d)$.Clearly $(X,\preceq)$ is an uncountable linear order. You can picture $(X, \preceq)$ as a $\omega_1$-long series of countable dense linear orders without endpoints.

Given any $(a,b) \in X$, the initial sequence $(\{(b,c) \in X \mid (b,c) \prec (a,b) \}, \preceq \restriction \{(b,c) \in X \mid (b,c) \prec (a,b) \})$ is a countable, dense linear order without a least element. Thus it's isomorphic to either $\mathbb Q \cap (0,1)$ or $\mathbb Q \cap (0,1]$.

Note that $\mathbb Q \cap (0,1)$ and $\mathbb Q$ are order isomorphic, so you can replace $\mathbb Q \cap (0,1)$ with $\mathbb Q$ in my above claim.

$\endgroup$
  • 2
    $\begingroup$ +1...An initial segment of $\omega_1 \times Q$ is order-dense, countable and has no end points so it is isomorphic to $Q$. (A result on linear orders due to Cantor.) $\endgroup$ – DanielWainfleet Feb 22 '16 at 3:02
  • 1
    $\begingroup$ @user254665 Actually, you are right. The way I've defined initial segments (by using $\prec$ instead of $\preceq$), there never is a maximal element. So any initial segment of this order is indeed order-isomorphic to $\mathbb Q$. $\endgroup$ – Stefan Mesken Feb 22 '16 at 3:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.