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 $P\_{2}$. 



Non-dense uncountable linear orderings
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.


