Equip $\mathbb{R}^2$ with the lexicographic order $(x_1,y_1) < (x_2,y_2)$ iff $x_1 < x_2$ or $x_1 = x_2 \Rightarrow y_1 < y_2$. Consider the induced order topology generated by the subbase consisting of intervals $](a,b), \infty[ := \{ (x,y) \ | \ (a,b) < (x,y) \}$ and $]-\infty, (x,y)[ := \{ (x,y) \ | \ (x,y) < (a,b) \}$ with $(a,b) \in \mathbb{R}^2$. The intersection of two of such intervals is an open ray $](a_1, b_1), (a_2, b_2)[$ with $(a_1, b_1) < (a_2, b_2)$. In particular, we see that the rays $](a,-\infty),(a,\infty)[$ ("parallel to the $y$-axis at $x = a$") are open as a union of $](x,-y),(x,y)[$ with $y$ ranging over the index set $[0, \infty)$. Thus, $\mathbb{R}^2$ with this topology is an uncountable disjoint union of real lines $\mathbb{R}$ with the usual order topology. It seems to be natural to say that $(\mathbb{R}^2, <)$ has then topological dimension $1$.
Question 1: Is there a precise useful definition of the dimension of a LOTS? I know only of a Hausdorff dimension but this needs a metric structure.
Question 2: Consider the subset $[0, \infty)^2$ with the induced order. Then it seems that the induced subspace topology differs from the order topology since sets like $[(a,0), (a,\infty)[$ are not open in the order topology of the induced lexicographic order on $[0, \infty)^2$. What is the "dimension" of such a space? It seems to be something like a "closed long ray" with lines $[(a,0), (a,\infty)[$ glued together So, it should have also dimension 1 in a specific sense.
They have a topological dimension, like ind, Ind or dim, all of which are defined for normal spaces, which LOTS are.
There is a theorem (a reference is in my answer here) that these dimensions coincide for all LOTS and that the only possible values are 0 or 1. So yes, dimension 1 will be the result, because dimension 0 for ordered spaces is equivalent to being totally disconnected.