# Hartogs' Lemma (Set Theory)/Proof 1

## Theorem

Let $S$ be a set.

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

## Proof

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 $\alpha \to S$.

$\blacksquare$

## Notes

• The construction in the proof in fact defines the least such ordinal, which is called the Hartogs number of $S$.
• Hartogs' lemma forms the basis for transfinite induction: There can be no strictly increasing mapping from the class of ordinals into a poset $S$, because if there were then it would be an injection from the Hartogs number of $S$ into $S$

## Source of Name

This entry was named for Friedrich Moritz Hartogs.