# Non-dense uncountable linear orderings

While I was working on my paper I came across this question:

$\mathbf{Question}$. Suppose $(X,<)$ is an uncountable linear ordering of cardinality $\kappa$. Is there a subset $Y$ of $X$ such that $|Y|=\kappa$ and for any two elements $y_{1},y_{2}$ of $Y$ with $y_{1}<y_{2}$ there is an $x\in X$ such that $y_{1}<x<y_{2}$?

I add that by a very simple argument using the Ramsey theorem I can give a positive answer to the above question when $\kappa$ is a weakly compact cardinal. By the same argument and this time using the Erdos-Rado partition theorem I can give a positive answer when $|X|=(2^{\mu})^{+}$ and $|Y|=\mu^{+}$ for any infinite cardinal $\mu$. Although this is enough for my purpose, the above question may remain interesting. I should indicate that I am not interested in any additional assumption on $(X,<)$.

Here is the argument for the case $\kappa$ being a weakly compact cardinal: We partition $[X]^{2}$ into two classes $P_{1},P_{2}$. Put an element $\{x_{1},x_{2}\}$ from $[X]^{2}$ in $P_{1}$ if there is no element of $X$ between $x_{1},x_{2}$, otherwise put it in $P_{2}$. By the Ramsey theorem for $\kappa$, there is a $Y\subset X$ such that $|Y|=\kappa$ and all elements of $[Y]^{2}$ lie in one partition class. It is easily seen that that partition class must be $P_{2}$.

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As a related fact: For any infinite cardinal $\kappa$, $\text{DLO}$ has an extremely dense model of size $\kappa$ that is a model with exactly $\kappa$-many points between any two distinct points. For example $\mathbb{R}$ is an extremely dense model of $\text{DLO}$ of size $2^{\aleph_0}$. To see this it suffices to take an arbitrary model of $\text{DLO}$ of size $\kappa$ like $M$ and fix a particular point $o\in M$ as origin then consider the following model of $\text{DLO}$: $\langle \{ f:\omega\rightarrow M ~|~\text{f is everywhere o except in finitely many points}\},<_{lex}\rangle$ –  Saint Georg Nov 6 '13 at 9:03

Define an equivalence relation $\sim$ on $X$ by $x \sim y$ if the interval between $x$ and $y$ is finite. Each $\sim$-equivalence class is either finite or countable. Therefore $|X/{\sim}| = \kappa$ since $\kappa$ is uncountable.
Any $Y \subseteq X$ that meets every ${\sim}$-class in exactly one point will be as required since the $X$-interval between any two points of $Y$ must be infinite.
There's a problem here. If $X$ is dense, then no intervals are finite, so every equivalence class is a singleton, in which case if $Y$ meets every equivalence class, $Y=X$. –  Asaf Karagila Nov 5 '13 at 17:21
@Asaf - That's not a problem: if $X$ is dense then $Y = X$ is a solution! –  François G. Dorais Nov 5 '13 at 17:22
@shahram: The quotient $X/{\sim}$ is often called the finite condensation of $X$. The idea goes back to Hausdorff (or even before) but a very thorough account is given in Rosenstein's Linear Orderings. –  François G. Dorais Nov 6 '13 at 6:32