Condition for Set Equivalent to Cardinal Number

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Theorem

Let $S$ be a set.

Let $\left\vert{S}\right\vert$ denote the cardinal number of $S$.

That is, let $\left\vert{S}\right\vert$ be the intersection of all ordinals equivalent to $S$.


Note that in the absence of the Axiom of Choice, $\left\vert{S}\right\vert$ may be the class of all sets.


Then the following are equivalent:

$(1): \quad S \sim \left\vert{S}\right\vert$
$(2): \quad \exists x \in \operatorname{On}: S \sim x$
$(3): \quad \exists x \in \operatorname{On}: \exists y: \left({y \subseteq x \land S \sim y}\right)$

where $A \sim B$ means that there is a bijection from $A$ onto $B$, and the quantification over $y$ is unbounded (so $y$ may be any set).


Proof

$2 \implies 1$

If $\exists x \in \operatorname{On}: S \sim x$, then by Ordinal Class is Strongly Well-Ordered by Subset there is a smallest ordinal $x$ such that $S \sim x$.

This smallest ordinal $x$ is the cardinal number of $S$, by definition.

$\Box$


$3 \implies 2$

Suppose that $y \subseteq x$ and $S \sim y$ for some ordinal $x$.

Since $y \subseteq x$, it follows that $y \sim z$ for some $z \in \operatorname{On}$ by Unique Isomorphism between Ordinal Subset and Unique Ordinal.

Therefore, by Set Equivalence is Equivalence Relation, $S \sim z$.

$\Box$


$1 \implies 3$

Suppose that $3$ is not true.

It follows that $S \not \sim x$ for any ordinal $x$.

\(\displaystyle \bigcap \left\{ {x \in \operatorname{On} : S \sim x}\right\}\) \(=\) \(\displaystyle \bigcap \varnothing\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \mathbb U\) $\quad$ by Intersection of Empty Set $\quad$

But $S \not \sim \mathbb U$, so $S \not \sim \left\vert{S}\right\vert$ by the definition of cardinal number.

The result follows by contraposition.

$\blacksquare$


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