Ordinal Less than Successor Aleph

Theorem
Let $x$ and $y$ be ordinals.

Then:


 * $y < \aleph_{x + 1} \iff y < \aleph_x \lor y \sim \aleph_x$

Sufficient Condition
But $\card y$ is a cardinal number, so it is either finite or an element of the infinite cardinal class.

It is a finite set $y \in \omega$ by Ordinal is Finite iff Natural Number.

If $y \in \omega$, then $y \in \aleph_x$ since $\aleph_x$ is an infinite set by Aleph is Infinite.

If it is an element of the infinite cardinal class, then $y = \aleph_z$ for some ordinal $z$ by the definition of the aleph mapping.

This means that:

Case 1: $z < x$
In the first case $z < x$:

Case 2: $z = x$
Therefore, $y < \aleph_x \lor y \sim \aleph_x$.

Necessary Condition
Suppose $y < \aleph_x \lor y \sim \aleph_x$.

Case 1: $y < \aleph_x$
Suppose $y < \aleph_x$.

Since $\aleph$ is strictly monotone by the definition of the aleph mapping:


 * $y < \aleph_{x + 1}$

Case 2: $y \sim \aleph_x$
Suppose $y \sim \aleph_x$.

Then by Equivalent Sets have Equal Cardinal Numbers:


 * $\card y = \aleph_x$

But since $\aleph$ is strictly monotone:


 * $\aleph_x < \aleph_{x + 1}$

Therefore, by Ordinal in Aleph iff Cardinal in Aleph:


 * $\card y < \aleph_{x + 1}$ and $y < \aleph_{x + 1}$