# Ordinal Less than Successor Aleph

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## Contents

## 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.