Natural Number m is Less than n iff m is an Element of n

Theorem

Let $\omega$ be the set of natural numbers defined as the von Neumann construction.

Let $m, n \in \omega$.

Then:

$m < n \iff m \in n$

That is, every natural number is the set of all smaller natural numbers.

Proof

By definition of the ordering on von Neumann construction:

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

Necessary Condition

Let $m \in n$.

Then from Natural Number is Transitive Set:

$m \subseteq n$

But if $m = n$, we would have:

$m \in m$

which contradicts Natural Number is Ordinary Set.

So we have:

$m \le n$ and $m \ne n$

that is:

$m < n$

$\Box$

Sufficient Condition

Let $m < n$.

Then by Natural Number m is Less than n implies n is not Greater than Successor of n:

$m^+ \le n$

But we have $m \in m^+$ by construction.

Hence:

$m \in n$

$\blacksquare$

Sources

• 2010: Raymond M. Smullyan and Melvin Fitting: Set Theory and the Continuum Problem (revised ed.) ... (previous) ... (next): Chapter $3$: The Natural Numbers: $\S 5$ Applications to natural numbers: Theorem $5.6$