# Zero topological dimension of ordered subspaces

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.

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 $$S$$.
• I added the last paragraph in case it isn't clear how to find an ordered set $S$ with a dense subset $D$ such that $|S|\gt|D|\ge|\mathbb R|$. – bof Aug 14 '20 at 20:00