# Natural Number Ordering is Transitive

## Theorem

Let $m, n, k \in \N$ where $\N$ is the set of natural numbers.

Let $<$ be the relation defined on $\N$ such that:

$m < n \iff m \in n$

where $\N$ is defined as the minimal infinite successor set $\omega$.

Then:

$k < m, m < n \implies k < n$

That is: $<$ is a transitive relation.

## Proof

Let $k < m, m < n$.

By definition it follows that $k \in m, m \in n$.

We have from Element of Finite Ordinal iff Subset that:

$k \in m \iff k \subseteq m$
$m \in n \iff m \subseteq n$

It follows from Subset Relation is Transitive that $k \subseteq n$.

Hence the result.

$\blacksquare$