Do all linear orders have Lebesgue dimension $\le 1$?

Question

It seems intuitively obvious that a linear order should have Lebesgue covering dimension $1$, but I've spent quite a bit of time and been unable to prove this. I would like to know if it is true, and how to prove it.

A 'creeping along' proof can be used for complete linear orders, and I thought for a while this would prove it for all orders, by embedding them, but it's possible for a space to have a subspace of higher dimension. Beyond that, I have no idea how to go about a proof.

I would also be interested to know if it is true for large and small inductive dimension.

Answer 1

In the paper "On the dimension of ordered spaces" by Bernard Brunet we have the following:

Definition. A topological space $X$ is called a line if it is homeomorphic to some subset of a totally ordered space.

Then the author proves:

Theorem 6.4 For any line $X$ we have $ind X=Ind X=dim X\in\{-1,0,1\}$, and moreover

1. $indX=-1$ if and only if $X=\emptyset$
2. $indX=0$ if and only if $X$ is totally disconnected
3. $indX=1$ if and only if $X$ is not totally disconnected

For the proof you need to read the paper. It requires some time, not straight forward.