Let's say that a linear order is $\omega_1$-short if it has no uncountable well-ordered or reverse well-ordered subset. For instance, linear orders satisfying Suslin's condition in his hypothesis are $\omega_1$-short. There are $\omega_1$-short linear orders which do not satisfy the countable chain condition.

Is there a $\omega_1$-short linear order in which every $\omega_1$-short linear order embeds?

Any $\omega_1$-short linear order embeds in the complete lexicographically ordered binary tree $2^{<\omega_1}$ (which can be ssen as a subfield of the ordered Field of surreals), but $2^{<\omega_1}$ is not $\omega_1$-short. I do not think that any of the $2^{<\alpha},\alpha < \omega_1$ qualify, even though they are $\omega_1$-short, but I don't have a proof. I actually don't know yet if $2^{<\alpha+1}$ can embed in $2^{< \alpha}$ for some values of $\alpha < \omega_1$.


1 Answer 1


The answer is no. See Corollary 5 in

Their argument is as follows: given an arbitrary linear order $L$, one can consider the set $\sigma L$ of all well ordered subsets of $L$ with end extension as a partial order. This partial order is extended to a linear order by the lexicographic ordering. It is easy to see that if $L$ is $\omega_1$-short then $\sigma L$ must be $\omega_1$-short as well.

However $\sigma L$ never embeds into $L$ (so $L$ is not universal). Indeed, if $f:\sigma L\to L$ is an order preserving embedding then define $s(\alpha)=f(s|\alpha)$ by induction on $\alpha \in ORD$ (taking unions at limit steps). If $\alpha<\beta$ then $s|\beta$ is an end extension of $s|\alpha$ so $s(\alpha)=f(s|\alpha)<f(s|\beta)=s(\beta)$. I.e. $s:\mathrm{ORD}\to L$ is injective which is clearly impossible.

  • $\begingroup$ This is a very nice argument (I like it when $Ord$ tries to inject inself into some sets); thank you for the reference as well. $\endgroup$
    – nombre
    Feb 7, 2017 at 10:22

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