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Paul McKenney
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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.

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?

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.

Source Link
Paul McKenney
  • 1.3k
  • 9
  • 14

Can a Suslin line be 2-entangled?

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?