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 $y < \aleph_{x + 1}$