# a partial order not dense iff a measurable exists

For $\kappa>0$ a regular cardinal, let $Ht_\kappa$ denote the following partial quasi-order:

(i) elements(objects) of $Ht_\kappa$ are classes X of sets of size $\kappa$ with the property that, ($<_\kappa$) for every $S\subseteq X$ of $|S| <\kappa$, there is $x_S\in X$ and $a_S, |a_S|<\kappa$, such that $\cup S\subseteq x_S\cup a_S$.

$X\leq_\kappa Y$ in $Ht_\kappa$ iff for every $x\in X$ exists $y\in Y$ such that $|x\setminus y|<\kappa$

To avoid set-theoretic difficulties, you may want instead to consider $Ht_\kappa(\lambda)$ ---the suborder consisting only of classes of subsets of $\lambda$---for some big cardinal $\lambda$.

It is easy to prove that $\kappa$ is measurable or countable iff $Ht_\kappa$ is not dense. (Proof : $I\leq_\kappa \{\kappa\}$ means that $I$ is a $\kappa$-closed ideal on $\kappa$; $I$ is maximal such iff there is nothing strictly $<_ \kappa$-between $I$ and $\{\kappa\}$. Finally, note that there is nothing between $X\leq_\kappa Y$ implies that for every $y\in Y$, there is nothing in between {$x\cap y:x\in X$ }$\leq_\kappa${ $y$} and $X<_\kappa Y$ implies that at least one of these inequalities is strict).

Is it consistent that $Ht_\omega$ is elementary equivalent to $Ht_\kappa$ for some $\kappa>\omega$ (in the language of partial orders)? What is this on the scale of large cardinals?
Does there exist a "canonical" monotone function $\mu:Ht_\kappa \longrightarrow \{0,1\}$ into the two-element order 0<1 ?
"Canonical" here means that for every automorphism $s: Ht_\kappa\longrightarrow Ht_\kappa$, $\mu(s(X))=\mu(X)$; "continious" means that for every set $X$, $\mu(\sup_i X_i)=\sup_i\mu(X_i)$ and $\mu(\inf_i X_i)=\inf_i\mu(X_i)$ (whenever these are well-defined).
If the latter is too strong, one may require the same only for directed $X$. (Note that $sup X$ and $inf X$ exist in $Ht_\kappa$ for every $X$). I can think of only functions which are cut-offs of $\Bbb Lcard (X)=\min$ { $|Y|: Y\geq X$}...