# 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.