So given a cardinal $\kappa$, i am trying to find a dense linear order with no end points of cardinality greater then $\kappa$ which has a dense subset of size $\kappa$
I think I'm almost there
What i did is to take some $M\vDash DLO$ of size $\kappa$ and then define $N$ to be the collection of subsets $U$ of $M$ such that $U$ has no greatest element and is closed from below, i.e if $x\in U$ and $y<x$ then $y\in U$ and with inclusion we can see that $(N,\subseteq )\vDash DLO$ and the map $x\mapsto \{y\mid y<x\}$ embed $M$ as a dense subset of $N$. Now i believe that $N$ is of size greater then $\kappa$ (maybe $2^\kappa$?) but i am unable to show it... help?
I want to show it so i can prove that $DLO$ is not $\kappa$-stable for all $\kappa$.
This won’t work, unfortunately: it’s possible that $|N|=\kappa$. (Consider, for example, the case in which $M$ is $\Bbb R$ with the usual order.)
HINT: Let $\lambda$ be the smallest cardinal such that $2^\lambda>\kappa$, and take the lexicographic order on ${^\lambda\Bbb Q}$, the set of functions from $\lambda$ to $\Bbb Q$. For your dense set look at functions of bounded support, where
$$\operatorname{supp}(f)=\{\xi<\lambda:f(\xi)\ne 0\}\;.$$