# Aleph-Null

## Theorem

Let $\omega$ denote the minimally inductive set.

$\omega = \aleph_0$

where $\aleph$ denotes the aleph mapping.

## Proof

For all $n \in \omega$, $n \notin \NN'$ by the definition of the class of infinite cardinals.

Therefore, $\omega \le \aleph_0$.

$\Box$

Moreover, $\omega \in \NN'$ by Minimally Inductive Set is Infinite Cardinal.

Therefore, $\aleph_x = \omega$ for some ordinal $x$.

It follows that $\aleph_0 \le \aleph_x$ since $0 \le x$ and the definition of the aleph mapping, so $\aleph_0 \le \omega$.

Thus, $\aleph_0 = \omega$.

$\blacksquare$