Zero topological dimension of ordered subspaces

Question

Following this thread, I was wondering whether the following generalized claim is true:

Given an ordered topological space $X$ with infinite cardinality $\kappa$ and a subset $A\subseteq X$ such that $\vert A\vert<\kappa$, then $A$ is zero dimensional.

This seems like it should be true, but I'm unsure how to prove it. Would an infinite ordinal space of cardinality greater than $\aleph_0$ be a counter example?

If the previous claim was not true would the following claim be true instead:

Given an ordered topological space $X$ with infinite cardinality $\kappa$ and a dense subset $A\subseteq X$ such that $\vert A\vert<\kappa$, then $A$ is zero dimensional.

Answer 1

Nope. Let $S$ be an ordered set with $|S|=\kappa\gt|\mathbb R|$ and $S\cap\mathbb R=\emptyset$. Let $X=\mathbb R\cup S$ be ordered so that every element of $\mathbb R$ precedes every element of $S$. Then $|\mathbb R|\lt\kappa=|X|$, and $\mathbb R$ has its usual topology, which is not zero-dimensional.

For the second question let us further suppose that $S$ has a dense subset $D$ such that $|S|\gt|D|\ge|\mathbb R|$, and let $A=\mathbb R\cup D$. Then $|A|=|D|\lt|S|=|X|$, and $A$ is dense in $X$ but not zero-dimensional because $\mathbb R$ is not zero-dimensional.

If $\lambda$ is the least cardinal such that $2^\lambda\gt2^{\aleph_0}$ then we can take for $S$ the set $\{0,1\}^\lambda$ ordered lexicographically, and for $D$ the set of all eventually constant functions in $S$.