# Total order isomorphic to $\mathbb N$

Suppose that we have a countable set $\mathcal C,$ and there is a total order $R$ on this set. By an analogy with the definition of the natural numbers, I guess that, if $R$ satisfy the following conditions, then $(\mathcal C,R)$ is isomorphic to $\mathbb N.$

$\bf 1)$ $\mathcal C$ has a minimum element with respect to $R$;

$\bf 2)$ $\forall\ x\in \mathcal C,$ there is a minimum element of the set $\left\{y\in\mathcal C|\ x<_R y\right\}$ with respect to $R.$

Question: Is my guess correct? If not, what improvement should me made to the requirements of $R$ ? Any hints are welcomed.

## 1 Answer

You're asking, practically, that $\cal C$ is a well-ordered set and without a maximum. There are uncountably many different ways to well-order a countable set, and uncountably many of them do not have a maximum either (and in a technical sense "almost all of them").

For example, consider the following ordering on $\Bbb Z$:

$$n\prec m\iff\begin{cases}0\leq n<m &\text{ or }\\m<n<0 & \text{ or }\\m<0\leq n\end{cases}$$

In other word, take all the negative integers, flip their order, and stack them on top of the natural numbers. So you effectively have two copies of $\Bbb N$ one after the other. It's easy to check that this satisfies the condition you impose.

However, you can find a condition that ensures a linear ordering is isomorphic to $\Bbb N$:

1. There is no maximal element.
2. Every proper initial segment is finite.

And I will let you prove this yourself, as a fun little exercise in order theory.

• I see what I've missed now. Thanks for your answer :). – painday Apr 12 '18 at 16:13