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.


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.

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    $\begingroup$ +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
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    $\begingroup$ @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

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