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A Suslin line is a linear order $L$ which is dense with no endpoints, complete, and ccc but not separable. I'm wondering what kind of order-preserving maps there are from $L$ into $L$. Specifically,

Question: Can there be a Suslin line $L$ such that for every one-to-one, monotonic function $f$ from an uncountable subset of $L$ into $L$, the set of $x\in\mathrm{dom}(f)$ with $f(x)\neq x$ is countable?

The motivation for this question is the following. A linear order $L$ is $n$-entangled if for every uncountable set $A$ of pairwise-disjoint $n$-tuples in $L$, and for every $s : n\to 2$, there are $a,b\in A$ such that $a_i < b_i$ if and only if $s(i) = 0$, for all $i < n$. One can show that a linear order $L$ is $2$-entangled if and only if it is rigid, in the sense that every one-to-one, monotonic function on an uncountable subset of $L$ is equal to the identity on a co-countable subset of its domain.

It's not difficult to show that a weakening of the Open Coloring Axiom implies there are no $2$-entangled sets of reals. However, OCA is consistent with the existence of a Suslin line. So an answer to the above question would provide evidence for an answer to the following.

Question: Is OCA consistent with the existence of a $2$-entangled linear order?

Edit: I forgot to mention why entangledness is relevant. A $3$-entangled linear order is necessarily separable, so $2$-entangledness is the most you might possibly get out of a Suslin line.

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Your property is too strong; there is no Suslin line like that.

The reason is that every Suslin line contains a copy of the real line, and we can define a counterexample function $f$ that concentrates only on this copy of $\mathbb{R}$. Specifically, suppose that $L$ is a Suslin line. Since the order is dense, we may find a copy of the rational line $\mathbb{Q}$ inside $L$, that is, a suborder $Q\subset L$ which is a countable dense endless order. Since $L$ is complete, we may add points to $Q$ realizing all the Dedekind cuts in $Q$, thereby extending $Q$ to a suborder $R\subset L$ that is order-isomorphic to the real line $\mathbb{R}$. Now, we may define $f$ on this copy of $\mathbb{R}$ to be the analogue of adding one, say, or any other order-preserving map with no fixed points. This gives an injective order-preserving map $f:R\to R$ for an uncountable subset $R\subset L$, with no fixed points.

Meanwhile, a positive answer is possible if one considers an analogue of your property on the underlying Suslin tree. Your property on the underlying Suslin trees is a kind of strong rigidity property of the kind considered in my article

In that article, we define that a Suslin tree $T$ has the unique branch property if forcing with $T$ necessarily adds only one branch through the tree. This is a strengthening of rigidity, since if $T$ has a nontrivial automorphism $\pi$, then one find a condition $x$ in $T$ with $\pi(x)\neq x$, and then forcing below $x$ will add two branches, namely, the generic branch $g$ and also $\pi[g]$, which will be different. We show that the Suslin trees obtained by the usual forcing to add a Suslin tree have the unique branch property, and also one can construct them in $L$ using the $\Diamond$ principle.

This is an analogue of your property for the trees:

Theorem. If $T$ is a Suslin tree with the unique branch property, then there is no function $f:S\to T$ with $S$ an uncountable set of extensions of a fixed node $x$, where $f(x)\perp x$.

Proof. Consider the uncountable set $S\subset T$. We may regard $S$ as a tree on its own, and in fact, $S$ is a Suslin tree. Force with $S$, so that we add an $\omega_1$-branch $b$ through $s$ in $V[b]$. It follows that the image $f[b]$ of this branch is a cofinal branch through $T$, and this is a different branch because $x\perp f(x)$. But note that $b$ is a cofinal branch through $T$, and since $T$ is Suslin, it follows that $b$ is also generic for forcing over $T$. Thus, in the forcing extension $V[b]$, we have two branches through $T$, which violates the unique branch property. QED

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  • $\begingroup$ I am still curious about the second question, however.. $\endgroup$ – Paul McKenney Jan 16 '14 at 0:25
  • $\begingroup$ In their paper "On the consistency of some partition theorems for continuous colorings and the structure of $\aleph_1$-dense real order type" Abraham, Rubin and Shelah prove the consistency of OCA and an increasing set (a somewhat analogue of a 2-entangled set). $\endgroup$ – Eran Jan 16 '14 at 20:51
  • $\begingroup$ @Eran, I would suggest that you post that as an answer, particularly if you can describe the issues a little more fully. $\endgroup$ – Joel David Hamkins Jan 17 '14 at 1:05
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Regarding the second question:

Definition: $A \subset \mathbb{R}$ of power $\aleph_1$ is called an increasing set if in every uncountable set of pairwise disjoint finite sequences from A there are two sequences $<a_i>$ and $<b_i>$ having the same length such that for all i $a_i<b_i$.

If $A$ is increasing then any strickly order reversing $f:A'\to A$ ($A'$ an uncountable subset of $A$) has a fixpoint.

In their paper "On the consistency of some partition theorems for continuous colorings and the structure of $\aleph_1$-dense real order types" Abraham, Rubin and Shelah prove the consistency of OCA and an increasing set (a somewhat analogue of a 2-entangled set as explained above). Note that Todorcevic' Open Coloring Axiom is a related but differently defined axiom than that of Abraham, Rubin and Shelah.

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  • $\begingroup$ Yes, I was referring to Todorcevic's OCA, which is notably different from the Abraham-Rubin-Shelah version. Thanks for your answer, though, it seems that I should probably dig back into their paper.. $\endgroup$ – Paul McKenney Jan 17 '14 at 19:24
  • $\begingroup$ I think you can say more about functions $f : A'\to A$ when $A$ is increasing; namely, $f(a) \ge a$ for all but countably-many $a\in A'$. $\endgroup$ – Paul McKenney Jan 17 '14 at 19:29

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