Hartogs' Lemma (Set Theory)

Theorem
Let $S$ be a set.

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

Proof
Define $\alpha = \left\{{\beta: \text{$\beta$ is an ordinal and there is an injection $\beta \to X$}}\right\}$.

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

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


 * $W = \left\{{\left({S', \preceq}\right): \text{$S' \subseteq S$ and $\preceq$ well-orders $S'$}}\right\}$

By Woset is Isomorphic to Unique Ordinal, each $w \in W$ corresponds to a unique ordinal $\beta_w$.

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


 * $\left\{{\beta: \exists w \in W: \beta = \beta_w}\right\}$

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

Then by Composite of Injections is Injection, $i \circ \iota: \gamma \to X$ 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 $\alpha \to X$.