Is it possible to construct an order on $\Bbb R^2$ that turns it into a set with the least-upper-bound property?

Question

What I already know:

Any ordered field that has the l.u.b.p. is isomorphic to $\Bbb R$.

However thing is a bit different in my question, I only request such an order that turns $\Bbb R^2$ into an ordered set with l.u.b.p., but not necessarily an ordered field.

Below are two of my failed attempts:

1).$(a,b)<(c,d)$ if $a<c$ or if $a=c$ but $b<d$. Counter-example: the left-half plane $(x,y),x\le 0$.

2). In polar coordinate, with $r\le 0$ and $\theta\in[0,2\pi)$, and let $(0,0)\leftrightarrow (r=0,\theta=0)$. Then $(r_1,\theta_1)<(r_2,\theta_2)$ if $r_1<r_2$ or if $r_1=r_2$ but $\theta_1<\theta_2$. Counter-example: the unit circle. $(2,0)$ is one of its upper bounds but it has no l.u.p. of course.

So I'm wondering is it possible for such an order to exist?

Ps: I want such an order to be "non-degenerate": it should be able to distinguish distinct elements, that's to say, if $A, B$ are distinct , then exactly one of $A<B$ and $B<A$ is true.

Answer 1

Well, $\Bbb R^2$ has the same cardinality as $\Bbb R,$ so if $f:\Bbb R^2\to\Bbb R$ is a bijection, we can define an order $\prec_f$ on $\Bbb R^2$ by $\vec x\prec_f\vec y$ iff $f(\vec x)<f(\vec y).$ In this fashion, we have ordered $\Bbb R^2,$ and $\langle\Bbb R^2,\prec_f\rangle$ has the least upper bound property.

We can't do this if we require a lexicographic-type of ordering. If we order $\Bbb R^2$ in a lexicographic way, then $\Bbb R^2$ is the union of uncountably-many disjoint open intervals and fails to have the least upper bound property (as none of those intervals has a least upper bound).

Added: In fact, we may turn $\Bbb R^2$ into a field with the least upper bound property by this method, though not with the usual operations. Instead, we define operations $\oplus$ and $\odot$ by $$\vec x\oplus\vec y:=f^{-1}\bigl(f(\vec x)+f(\vec y)\bigr)$$ and $$\vec x\odot\vec y:=f^{-1}\bigl(f(\vec x)\cdot f(\vec y)\bigr).$$ Readily, the $\oplus$ identity is $f^{-1}(0),$ the $\odot$ identity is $f^{-1}(1),$ and we can verify all the field axioms are satisfied. It is fairly straightforward to prove that the usual addition operation on $\Bbb R^2$ does not agree with $\oplus.$

Answer 2

Quick, dirty, and probably useless answer: any well-ordering.

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More specifically, any non-empty set (in particular, $\mathbb R^2$) can be endowed with an ordering, called a well-ordering, $\succsim$ such that:

• $\mathbf x\succsim \mathbf x$ (for any $\mathbf x\in\mathbb R^2$);

• if $\mathbf x\succsim \mathbf y$ and $\mathbf y\succsim \mathbf x$, then $\mathbf x=\mathbf y$ (for any $\mathbf x,\mathbf y\in\mathbb R^2$);

• if $\mathbf x \succsim \mathbf y$ and $\mathbf y \succsim \mathbf z$, then $\mathbf x \succsim \mathbf z$ (for any $\mathbf x, \mathbf y,\mathbf z\in\mathbb R^2$);

• if $\mathbf x\neq \mathbf y$, then either $\mathbf x\succsim \mathbf y$ and $\mathbf y\succsim\mathbf x$, but not both (for any $\mathbf x,\mathbf y\in\mathbb R^2$); and

• if $X\subseteq \mathbb R^2$ and $X$ is not empty, then $X$ has a least element: there is some $\underline{\mathbf x}\in X$ such that $\mathbf x\succsim\underline{\mathbf x}$ for all $\mathbf x\in X$.

One can use this last condition to establish the least-upper-bound property: Fix any non-empty subset $S\subseteq\mathbb R^2$ and suppose it has an upper bound with respect to $\succsim$ (that is, there exists some $\mathbf x\in\mathbb R^2$ such that $\mathbf x\succsim\mathbf s$ for all $\mathbf s\in S$). Let $X$ denote the set of all upper bounds of $S$: $$X\equiv\{\mathbf x\in\mathbb R^2\,|\,\mathbf x\succsim\mathbf s\text{ for all $\mathbf s\in S$}\}.$$ Then, by assumption, $X\neq\varnothing$, so that $X$ has a least element, which is by definition the least upper bound of $S$.

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I said this answer was useless—this is because a notorious property of well-orderings is that they exist, but they are practically impossible to actually construct. In fact, the existence of well-orderings on arbitrary non-empty sets is such a deep set-theoretic result as to be equivalent to the axiom of choice.