Aleph is Infinite

Theorem
Let $x$ be an ordinal.


 * $\aleph_x \ge \omega$

where:
 * $\aleph$ denotes the aleph mapping
 * $\omega$ denotes the minimal infinite successor set.

Proof
Since $0 \le x$, it follows that $\aleph_0 \le \aleph_x$ by definition of the aleph mapping.

But $\aleph_0 = \omega$ by Aleph-Null.

Therefore, $\omega \le \aleph_x$.