Aleph-Null

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Theorem

Let $\omega$ denote the minimal infinite successor set.

$\omega = \aleph_0$

where $\aleph$ denotes the aleph mapping.


Proof

For all $n \in \omega$, $n \notin \mathcal N'$ by the definition of the infinite cardinal class.

Therefore, $\omega \le \aleph_0$.

$\Box$


Moreover, $\omega \in \mathcal N'$ by Minimal Infinite Successor 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, $\aleph_0 = \omega$.

$\blacksquare$