Aleph is Infinite
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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, $\omega \le \aleph_x$.
$\blacksquare$