Hartogs' Theorem

Theorem
Let $x$ be a set.

Then there exists an ordinal $n$ such that there does not exist an injection from $n$ to $x$.

That is:
 * $\left\vert{n}\right\vert \npreceq \left\vert{x}\right\vert$

where $\left\vert{\cdot}\right\vert$ represents cardinality.

Proof
Let $w$ be the set of all well-orderings on subsets of $x$.

By Woset is Isomorphic to Unique Ordinal, there exists a mapping $F: w \to \operatorname{On}$ defined by letting $F \left({a}\right)$ be the ordinal which is isomorphic to $a$.

By Mapping from Set to Ordinal Class is Bounded Above, $F \left({w}\right)$ has an upper bound $m$.

Then if $n$ is any ordinal strictly greater than $m$, then $n \notin F \left({w}\right)$.

Seeking a contradiction, suppose that there is an injection $g: n \to x$.

Then by Injection to Image is Bijection, there is a bijection from $n$ onto $g \left({n}\right)$.

But this induces a well-ordering on $g \left({n}\right) \subseteq x$ which is isomorphic to $n$, contradicting the fact that $n \notin F \left({w}\right)$.