"$\mathbb{R}^2$ can't be totally ordered" (nicely)

Question

I have often heard (both online and in person) people say that "$\mathbb{R}^2$ can't be totally ordered." I would like to understand this statement.

Of course, on the face of it, this is false: Pick your favorite bijection $f:\mathbb{R}^2 \to \mathbb{R}$ and define $x \leq y$ iff $f(x) \leq f(y)$.

When I bring this up, people usually dismiss it, saying it isn't "nice" enough. This is fair, but now leaves me with the question of what a "nice" ordering would look like.

Other questions on this site (like this) show that there is no ordering which makes $\mathbb{C}$ an ordered field. I find this answer somewhat unsatisfying. I don't need to appeal to the algebraic structure of $\mathbb{R}$ to give it a "nice" ordering. Furthermore, I would like to be able to extend this notion of a "nice ordering" to other topological spaces that don't admit field structures: does $\mathbb{R}^3$ have a "nice" ordering? how about $S^1$?

Here's a definition I came up with: a total ordering on a topological space $X$ is "nice" if for every $x < y$, there are neighborhoods $U_x \ni x$ and $U_y \ni y$ so that for all $a \in U_x$ and $b \in U_y$, we have $a < b$.

So the usual ordering on $\mathbb{R}$ is "nice," but (for all $f$ I can think of) the ordering of $\mathbb{R}^2$ given above isn't.

I've tried proving that $\mathbb{R}^2$ and $S^1$ can't be given a nice total ordering under this definition, but have had some difficulty.

Questions:

1. Is there an established notion of a "nice" ordering on a topological space?

2. How can you prove that $\mathbb{R}^2$ (or $S^1$) can't be totally ordered nicely? (either with my definition or someone else's, if it exists)

Answer 1

I have not seen a standard definition of a nice total ordering, but I like the one you provided.

Neither $\mathbb{R}^2$ nor $S^1$ admit a nice ordering by your definition. The proof is straightforward:

Claim 1: In a topological space $X$ with a nice total ordering $<$, for all $x\in X$ the sets $\{y \mid y<x\}$ and $\{y \mid y> x\}$ are open.

Claim 2: For $X = \mathbb{R}^2$ or $X=S^1$, there is no way to write $X=A\cup B\cup\{x\}$ for disjoint nonempty open sets $A,B$.

The proof of each claim is straightforward: For the first, note that the set $\{y\mid y<x\}$ can be written as a union of open balls around all points inside itself, and similarly for $\{y \mid y>x\}$. For the second, observe that the compliment of a singleton set in $\mathbb{R}^2$ or in $S^1$ is connected.

Answer 2

Suppose you have a point $x_0 \in X$ such that $X \setminus \{ x_0 \}$ is connected. Then under your definition of a "nice" total order, $\{ x \in X \mid x < x_0 \}$ and $\{ x \in X \mid x_0 < x \}$ will be a partition of $X \setminus \{ x_0 \}$ into two disjoint open subsets; therefore, one of them must be the whole subspace $X \setminus \{ x_0 \}$. In other words, $x_0$ must be either a maximum or a minimum element of $X$ under the given order.

Now, if you can find at least three distinct points $x_0, x_1, x_2$ such that $X \setminus \{ x_i \}$ is connected for each $i \in \{ 0, 1, 2 \}$, then you will get a contradiction. This is the case both for $X = \mathbb{R}^2$ and for $X = S^1$.

Answer 3

There is a notion of a GO-space (generalised ordered space), i.e. a topological space $(X,\tau)$ such that $X$ has a linear order such that the order topology induced by $<$ is a subset of $\tau$ and $\tau$ has a base of order-convex sets ($A \subseteq X$ is order convex iff $\forall x,y \in A:\forall z \in X: x \le z \le y \to z \in A$).

All LOTS (linearly ordered topolgoical spaces) $X$ (where $X$ has the order topology w.r.t. some linear order $<$) are GO-spaces as the order-topology base (of segments and intervals) is order convex. The Sorgenfrey line ($\Bbb R$ in the topology with base $\{[a,b)\mid a < b \in \Bbb R\}$) is a standard example of a GO-space (that cannot be a LOTS, by deeper results).

$\Bbb R^2$ cannot be a GO-space (which I take as an interpretation of "can be nicely ordered topologically") because in a connected GO-space, all but two of its points are cutpoints (removing them makes the remainder disconnected), or because $\dim(X) \le 1$ for all GO-spaces, and $\dim(\Bbb R^2)=2$ and the plane has no cutpoints. $S^1$ also has no cutpoints, so is no GO-space either.