User:Dfeuer/Element of Natural Number is Natural Number

Theorem

Let $n$ be a natural number.

Let $x \in n$.

Then $x$ is also a natural number.

Proof

Let $b$ be the set of all natural numbers $n$ such that $n \subseteq \omega$.

$0 \subseteq \omega$, so $0 \in b$.

Suppose that $n \in b$.

Let $p \in n^+$.

Then $p \in n$ or $p = n$.

If $p \in n$ then $p \in \omega$ because $n \in b$.

If $p = n$ then $p \in \omega$.

Since this holds for all $p \in n^+$, $n^+ \subseteq \omega$, so $n^+ \in b$.

We have shown that $n \in b \implies n^+ \in b$.

Thus $b$ is an inductive set, so by the definition of natural number, every natural number is in $b$.

Thus every element of every natural number is an element of $\omega$.

$\blacksquare$