Natural Numbers are Comparable/Strong Result

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Theorem

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 1

Let $\N$ be defined as the von Neumann construction $\omega$.

By definition of the ordering on von Neumann construction:

$m \le n \iff m \subseteq n$

From Von Neumann Construction of Natural Numbers is Minimally Inductive, $\omega$ is minimally inductive class under the successor mapping.

Then from Minimally Inductive Class under Progressing Mapping induces Nest, $\omega$ is a nest in which:

$\forall m, n \in \omega: \map g x \subseteq y \lor y \subseteq x$

From the definition of $\map g x$ in this context:

$\forall x \in \omega: \map g x = x^+$

That is:

$\forall m, n \in \N: m + 1 \subseteq n \lor n \subseteq m$

whence the result.

$\blacksquare$


Proof 2




Sources