Condition for Set Equivalent to Cardinal Number

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Theorem

Let $S$ be a set.

Let $\card S$ denote the cardinal number of $S$.

That is, let $\card S$ be the intersection of all ordinals equivalent to $S$.


Note that in the absence of the Axiom of Choice, $\card S$ may be the class of all sets.


Then the following are equivalent:

$(1): \quad S \sim \card S$
$(2): \quad \exists x \in \On: S \sim x$
$(3): \quad \exists x \in \On: \exists y: \paren {y \subseteq x \land S \sim y}$

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 $\exists x \in \On: S \sim x$, 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 $z \in \On$ 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 $x$.

\(\displaystyle \bigcap \set {x \in \On : S \sim x}\) \(=\) \(\displaystyle \bigcap \O\)
\(\displaystyle \) \(=\) \(\displaystyle \mathbb U\) by Intersection of Empty Set

But $S \not \sim \mathbb U$, so $S \not \sim \card S$ by the definition of cardinal number.

The result follows by contraposition.

$\blacksquare$


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