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

\(\ds y\) \(<\) \(\ds \aleph_{x + 1}\)
\(\ds \leadsto \ \ \) \(\ds \card y\) \(<\) \(\ds \aleph_{x + 1}\) Ordinal in Aleph iff Cardinal in Aleph

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


It is a finite set if and only if $y \in \omega$ 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:

\(\ds \aleph_z\) \(<\) \(\ds \aleph_{x + 1}\) by hypothesis
\(\ds \leadsto \ \ \) \(\ds z\) \(<\) \(\ds x + 1\) since $\aleph$ is an order isomorphism
\(\ds \leadsto \ \ \) \(\ds z < x\) \(\lor\) \(\ds z = x\) Definition of Successor Set


Case 1: $z < x$

In the first case $z < x$:

\(\ds z\) \(<\) \(\ds x\)
\(\ds \leadsto \ \ \) \(\ds \aleph_z\) \(<\) \(\ds \aleph_x\) Definition of Aleph Mapping
\(\ds \leadsto \ \ \) \(\ds \card y\) \(<\) \(\ds \aleph_x\) Definition of $\aleph_z$
\(\ds \leadsto \ \ \) \(\ds y\) \(<\) \(\ds \aleph_x\) Ordinal in Aleph iff Cardinal in Aleph


Case 2: $z = x$

\(\ds z\) \(=\) \(\ds x\)
\(\ds \leadsto \ \ \) \(\ds \aleph_z\) \(=\) \(\ds \aleph_x\) Substitutivity of Class Equality
\(\ds \leadsto \ \ \) \(\ds y\) \(\sim\) \(\ds \aleph_x\) Definition of $\aleph_z$


Therefore:

$y < \aleph_x \lor y \sim \aleph_x$

$\Box$


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 $y < \aleph_{x + 1}$
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