Is it possible to have a linear order that is not "on a line"?

Question

I am looking at some problems on linear order. It seems in all the problems, I am dealing with things that are 1D

Whether it is $\mathbb{R}$ itself, or $\left\{\dfrac{1}{n}|n \in \mathbb{Z}_+\right\}\;$ or $\;\mathbb{Q}$.

When I think of orders on the plane, the first thing that comes to my mind is the lexicographic/dictionary order, which is not a linear order.

So my question is whether there are linear orders on sets that are higher dimensional, or not on a line.

Answer 1

Please look at your question's comments to start.

To go in a slightly different direction, you might want to consider the well-ordering theorem which states that every set can be well-ordered. This immediately implies that $\mathbb{R}$ can be well-ordered. And, surprisingly (or not so), no well-order on $\mathbb{R}$ has been explicitly given. So, if a well-order (which is linear) on $\mathbb{R}$ exists, it is probably extremely complicated.

You're comment about 'not on a line' and dimensionality does not make a lot of sense. In its most abstract form, a linear order has nothing to do with a line. However, it is often helpful to think about it as such, since it is easy to grasp.

One such example of a set without a 'line' is to consider $$\{a,b,c\}$$ and its power set under inclusion.

Answer 2

Note that a linear order might be "too big" to fit in the reals. For instance, the Hartogs ordinal of the reals is a linear order (indeed, a well-order) but it cannot be embedded in the reals, so in that sense it doesn't "fit on the line".