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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.

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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 $S$.

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  • $\begingroup$ Just a question about the last paragraph, is it a way to construct any such examples as you wrote above? $\endgroup$ – Keen-ameteur Aug 14 '20 at 13:54
  • $\begingroup$ 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 $|S|\gt|D|\ge|\mathbb R|$. $\endgroup$ – bof Aug 14 '20 at 20:00

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