# Hartogs number

In mathematics, specifically in axiomatic set theory, a Hartogs number is a particular kind of cardinal number. It was shown by Friedrich Hartogs in 1915, from ZF alone (that is, without using the axiom of choice), that there is a least well-ordered cardinal greater than a given well-ordered cardinal.

To define the Hartogs number of a set it is not in fact necessary that the set be well-orderable: If X is any set, then the Hartogs number of X is the least ordinal α such that there is no injection from α into X. If X cannot be well-ordered, then we can no longer say that this α is the least well-ordered cardinal greater than the cardinality of X, but it remains the least well-ordered cardinal not less than or equal to the cardinality of X. The map taking X to α is sometimes called Hartogs' function.

## Proof

Given some basic theorems of set theory, the proof is simple. Let ${\displaystyle \alpha =\{\beta \in {\textrm {Ord}}\mid \exists i:\beta \hookrightarrow X\}}$, the class of all ordinal numbers β for which an injective function exists from β into X.

First, we verify that α is a set.

1. X × X is a set, as can be seen in axiom of power set.
2. The power set of X × X is a set, by the axiom of power set.
3. The class W of all reflexive well-orderings of subsets of X is a definable subclass of the preceding set, so it is a set by the axiom schema of separation.
4. The class of all order types of well-orderings in W is a set by the axiom schema of replacement, as
(Domain(w), w) ${\displaystyle \cong }$ (β, ≤)
can be described by a simple formula.

But this last set is exactly α.

Now because a transitive set of ordinals is again an ordinal, α is an ordinal. Furthermore, there is no injection from α into X, because if there were, then we would get the contradiction that αα. And finally, α is the least such ordinal with no injection into X. This is true because, since α is an ordinal, for any β < α, βα so there is an injection from β into X.