Posted by Jason Polak on 11. January 2018 · Write a comment · Categories: analysis · Tags: ,

A total ordering $\lt$ on a set $S$ is called dense if for every two $x,y\in S$ with $x \lt y$, there exists a $z\in S$ such that $x\lt z\lt y$. A total ordering is said to be without endpoints if for every $x\in S$ there exists $y,z\in S$ such that $y \lt x\lt z$. In other words, a total order is without endpoints if there is no largest element and no smallest element.

Obviously, if a total order is dense or without endpoints, it must be a relation on an infinite set. One such example is the rational numbers with the usual order.

Theorem. [Cantor] Up to order isomorphism, there is exactly one countable dense totally ordered set without endpoints.

So the rational numbers is the only example of such a totally ordered set up to isomorphism. In particular, this means that $\Q$ and $\Q – \{0\}$ must be order isomorphic. Can you find the isomorphism?

If we take Cantor's theorem and remove the word 'countable', then the new statement is no longer true! For example, Suppose we take the real numbers $\R$. Then we claim that $\R$ is not order isomorphic to $\R – \{0\}$.

Indeed, suppose $\varphi:\R\to\R-\{0\}$ is an order isomorphism. Consider the sequence $a_0,a_1,\dots$ defined by $\varphi(a_i) = 2^{-i}$. Since $\varphi(a_i) \gt \varphi(0)$, we see that $a_0,a_1,\dots$ is a bounded strictly decreasing sequence with limit $\lim a_i = a$.

Because $\varphi$ is an order isomorphism, we conclude $0 \gt\varphi(a)$. Now let $z\in \R$ be such that $0 \gt \varphi(z)\gt \varphi(a)$. Then $z\gt a$ and so $z \gt a_n$ for some $n$. But this means that $\varphi(z) \gt 0$, which is a contradiction! Hence no order isomorphism $\R\to \R-\{0\}$ exists.

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