Hartogs' Lemma (Set Theory)/Proof 2

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Theorem

Let $S$ be a set.

Then there exists an ordinal $\alpha$ such that there is no injection from $\alpha$ to $S$.

Proof

Let $W$ be the set of all well-orderings on subsets of $S$.

By the Counting Theorem, there exists a mapping $F: W \to \On$ defined by letting $\map F s$ be the ordinal which is isomorphic to $s$.

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By Mapping from Set to Class of All Ordinals is Bounded Above, $F \sqbrk W$ has an upper bound $\alpha_0$.

Then if $\alpha$ is any ordinal strictly greater than $\alpha_0$, then $\alpha \notin F \sqbrk W$.

Aiming for a contradiction, suppose there is an injection $g: \alpha \to S$.

Then by Injection to Image is Bijection, there is a bijection from $\alpha$ onto $g \sqbrk \alpha$.

But this induces a well-ordering on $g \sqbrk \alpha \subseteq S$ which is isomorphic to $\alpha$, contradicting the fact that $\alpha \notin F \sqbrk W$.

$\blacksquare$

Source of Name

This entry was named for Friedrich Moritz Hartogs.

Sources

• 2010: Raymond M. Smullyan and Melvin Fitting: Set Theory and the Continuum Problem (revised ed.): Theorem $9.5.1$