Natural Numbers are Comparable

Theorem
Let $\N$ be the natural numbers, defined as the minimal infinite successor set $\omega$.

Let $m, n \in \N$.

Then exactly one of the following is the case:
 * $(1): \quad m \in n$
 * $(2): \quad m = n$
 * $(3): \quad n \in m$

That is, two natural numbers are always comparable by the ordering $\le$ where:
 * $m \le n \iff \begin{cases}

m = n & \text{or}\\ m \in n & \end{cases}$

Proof
Proof by induction:

For each $n \in \omega$, let $S \left({n}\right)$ be the set of all $m \in \omega$ which are comparable with $n$.

Let $S$ be the set of all $n \in \N$ for which $S \left({n}\right) = \omega$.

Thus the proof is equivalent to demonstrating that $S = \omega$.

Basis for the Induction
First consider $S \left({0}\right)$.

Clearly $0 \in S \left({0}\right)$ as $0 = 0$.

Let $m \in S \left({0}\right)$.

Then $m \notin 0$ as $0 = \varnothing$.

So either:
 * $(1): \quad 0 = m$, in which case $0 \in m^+$ by definition of successor set

or:
 * $(2): \quad 0 \in m$, in which case, because $m \in m^+$ by definition of successor set, again $0 \in m^+$.

In all cases, $m \in S \left({0}\right) \implies m^+ \in S \left({0}\right)$.

So $S \left({0}\right) = \omega$ by induction.

This is the basis for the induction.

Induction Hypothesis
The induction hypothesis is that:
 * $S \left({k}\right) = \omega$

for some $k \in \omega$.

It is now necessary to show that it follows that:
 * $S \left({k^+}\right) = \omega$

Induction Step
This is the induction step:

Consider the set $S \left({k^+}\right)$, given that $S \left({k}\right) = \omega$.

From the basis for the induction, we have that $k^+ \in S \left({0}\right)$.

That is, $k^+$ is comparable with $0$.

So $0$ is comparable with $k^+$ and so $0 \in S \left({k^+}\right)$.

Suppose $m \in S \left({k^+}\right)$.

Then either:
 * $(1): \quad k^+ \in m$, in which case $k^+ \in m^+$

or:
 * $(2): \quad k^+ = m$, in which case the same applies: $k^+ \in m^+$

or:
 * $(3): \quad m \in k^+$.

In case $(3)$, either:
 * $(3 a): \quad m = k$, in which case $m^+ = k^+$

or:
 * $(3 b): \quad m \in k$.

Case $(3 b)$ is treated as follows.

We have that $m^+ \in S \left({k}\right)$ by the induction hypothesis.

Therefore, either:
 * $(4 a): \quad k \in m^+$

or:
 * $(4 b): \quad k = m^+$

or:
 * $(4 c): \quad m^+ \in k$.

$(4 a)$ is not compatible with $m \in k$, because either:
 * $k \in m$

or
 * $k = m$

and so $k \subseteq m$ which contradicts Natural Number is Not Subset of Element.

Both $(4 b)$ and $(4 c)$ imply that $m^+ \in n^+$.

Hence $S \left({k^+}\right) = \omega$.

It follows by the Principle of Finite Induction that $S = \omega$.

It then follows from Natural Number is Not Subset of Element that it is not possible for more than one of:
 * $(1): \quad m \in n$
 * $(2): \quad m = n$
 * $(3): \quad n \in m$

to be true at the same time.