Ordinal Less than Successor Aleph

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Theorem

Let $x$ and $y$ be ordinals.


Then:

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


Proof

Sufficient Condition

\(\displaystyle y\) \(<\) \(\displaystyle \aleph_{x+1}\)
\(\displaystyle \iff \ \ \) \(\displaystyle \left\vert{ y }\right\vert\) \(<\) \(\displaystyle \aleph_{x+1}\) by Ordinal in Aleph iff Cardinal in Aleph

But $\left|{ y }\right|$ is a cardinal number, so it is either finite or an element of the infinite cardinal class.


It is a finite set iff $y \in \omega$ 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:

\(\displaystyle \aleph_z\) \(<\) \(\displaystyle \aleph_{x+1}\) Hypothesis
\(\displaystyle \implies \ \ \) \(\displaystyle z\) \(<\) \(\displaystyle x+1\) since $\aleph$ is an order isomorphism
\(\displaystyle \implies \ \ \) \(\displaystyle z < x\) \(\lor\) \(\displaystyle z = x\) definition of successor


Case 1: $z < x$

In the first case $z < x$:

\(\displaystyle z\) \(<\) \(\displaystyle x\)
\(\displaystyle \implies \ \ \) \(\displaystyle \aleph_z\) \(<\) \(\displaystyle \aleph_x\) definition of aleph mapping
\(\displaystyle \implies \ \ \) \(\displaystyle \left\vert{ y }\right\vert\) \(<\) \(\displaystyle \aleph_x\) definition of $\aleph_z$
\(\displaystyle \implies \ \ \) \(\displaystyle y\) \(<\) \(\displaystyle \aleph_x\) by Ordinal in Aleph iff Cardinal in Aleph


Case 2: $z = x$

\(\displaystyle z\) \(=\) \(\displaystyle x\)
\(\displaystyle \implies \ \ \) \(\displaystyle \aleph_z\) \(=\) \(\displaystyle \aleph_x\) by Substitutivity of Class Equality
\(\displaystyle \implies \ \ \) \(\displaystyle y\) \(\sim\) \(\displaystyle \aleph_x\) by definition of $\aleph_z$


Therefore, $y < \aleph_x \lor y \sim \aleph_x$.

$\Box$


Necessary Condition

Suppose that $y < \aleph_x \lor y \sim \aleph_x$.


Case 1: $y < \aleph_x$

Suppose that $y < \aleph_x$.

It follows that $y < \aleph_{x+1}$ since $\aleph$ is strictly monotone by the definition of the aleph mapping.


Case 2: $y \sim \aleph_x$

Suppose that $y \sim \aleph_x$.

Then $\left|{ y }\right| = \aleph_x$ by Equivalent Sets have Equal Cardinal Numbers.


But $\aleph_x < \aleph_{x+1}$ since $\aleph$ is strictly monotone.

Therefore, $\left|{ y }\right| < \aleph_{x+1}$ and $y < \aleph_{x+1}$ by Ordinal in Aleph iff Cardinal in Aleph.