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 Isomorphic to Unique Ordinal, there is 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 Restricting Injection to 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 $n \notin F\left({ w }\right)$.