Aleph is Infinite

Theorem

Let $x$ be an ordinal.

$\aleph_x \ge \omega$

where:

$\aleph$ denotes the aleph mapping

$\omega$ denotes the minimally inductive 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, $\omega \le \aleph_x$.

$\blacksquare$