# Hartogs' Lemma (Set Theory)

*This proof is about Hartogs' Lemma in the context of Set Theory. For other uses, see Hartogs' Lemma.*

## Theorem

Let $S$ be a set.

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

## Proof 1

Define $\alpha = \set {\beta: \text{$\beta$ is an ordinal and there is an injection $\beta \to S$}}$.

First of all, it is to be shown that $\alpha$ is a set.

To this end, define the set $W$ by:

- $W = \set {\paren {S', \preceq}: \text{$S' \subseteq S$ and $\preceq$ well-orders $S'$}}$

By the Counting Theorem, each $w \in W$ corresponds to a unique ordinal $\beta_w$.

Thus by the Axiom of Replacement, the following is a set:

- $\set {\beta: \exists w \in W: \beta = \beta_w}$

It follows from Injection Induces Well-Ordering that this set coincides with $\alpha$.

In particular, then, $\alpha$ is a set.

Next, to establish $\alpha$ is an ordinal.

Suppose $\beta \in \alpha$ and $\gamma < \beta$.

Let $i: \beta \to S$ be an injection, and let $\iota: \gamma \to \beta$ be the inclusion of $\gamma$ in $\beta$.

Then by Composite of Injections is Injection, $i \circ \iota: \gamma \to S$ is an injection.

Hence $\gamma \in \alpha$, and therefore $\alpha$ is an ordinal.

Finally, by Ordinal is not Element of Itself, $\alpha \notin \alpha$.

That is to say, there is no injection $\alpha \to S$.

$\blacksquare$

## Proof 2

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 $\alpha \notin F \sqbrk W$.

$\blacksquare$

## Also known as

This result is also known as **Hartogs' theorem**.

## Source of Name

This entry was named for Friedrich Moritz Hartogs.