Natural Numbers are Comparable

Theorem

Let $\N$ be the natural numbers.

Let $m, n \in \N$.

Then $m$ and $n$ are comparable by the ordering relation $\le$.

That is, either:

$(1): \quad m \le n$

or:

$(2): \quad n \le m$

or possibly both.

Strong Result

Let $\N$ be the natural numbers.

Let $m, n \in \N$.

Then either:

$(1): \quad m + 1 \le n$

or:

$(2): \quad n \le m$

Proof

Let $\N$ be defined as the minimally inductive set $\omega$.

By definition of the ordering on minimally inductive set:

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

Thus it is sufficient to prove that exactly one of the following is the case:

$(1): \quad m \in n$

$(2): \quad m = n$

$(3): \quad n \in m$

The proof operates by induction:

For each $n \in \omega$, let $\map S n$ 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 $\map S n = \omega$.

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

Basis for the Induction

First consider $\map S 0$.

Clearly $0 \in \map S 0$ as $0 = 0$.

Let $m \in \map S 0$.

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

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, $m \in \map S 0 \implies m^+ \in \map S 0$.

So $\map S 0 = \omega$ by induction.

This is the basis for the induction.

Induction Hypothesis

The induction hypothesis is that:

$\map S k = \omega$

for some $k \in \omega$.

It is now necessary to show that it follows that:

$\map S {k^+} = \omega$

Induction Step

This is the induction step:

Consider the set $\map S {k^+}$, given that $\map S k = \omega$.

From the basis for the induction, we have that $k^+ \in \map S 0$.

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

So $0$ is comparable with $k^+$ and so $0 \in \map S {k^+}$.

Suppose $m \in \map S {k^+}$.

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:

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

or:

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

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

We have that $m^+ \in \map S k$ by the induction hypothesis.

Therefore, either:

$(4 \text a): \quad k \in m^+$

or:

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

or:

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

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

$k \in m$

or

$k = m$

and so $k \subseteq m$ which contradicts Finite Ordinal is not Subset of one of its Elements.

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

Hence $\map S {k^+} = \omega$.

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

$\Box$

It then follows from Finite Ordinal is not Subset of one of its Elements 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.

$\blacksquare$