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Every injection from an ordinal into a set $X$ has domain strictly below the Hartogs number $h(X)$. Therefore all partial injections with ordinal domain and range in $X$ form a set of graphs contained in $\mathcal{P}(h(X)\times X)$. Under extension, the union of a chain remains a partial injection of this kind. Zorn's lemma then supplies a maximal one, and extending at an unused range element shows that it must be a bijection onto $X$. This gives the well-ordering theorem with the set-sizedness of the poset made explicit.

For infinite sets in ZFC, the Hartogs number is the successor of the input cardinal. Every smaller ordinal injects into the input, whereas that successor does not. An infinite successor cardinal ${\kappa }^{+}$ is regular: a cofinal family of at most $\kappa$ ordinals below it would have union of cardinal at most $\kappa \cdot \kappa =\kappa$, a contradiction. Finite inputs give finite successor cardinals, so the formulation in terms of alephs requires an infinite input. The general Hartogs theorem itself does not require choice.

For every set $X$ there is a least ordinal $h(X)$ that does not inject into $X$.

The least ordinal supplied by the theorem is the Hartogs number.

Use ZFC and infinite input sets, the usual scope of the stated claim. For a finite set of size $n$, its Hartogs number is $n+1$, which is a finite successor cardinal rather than an aleph. Without the axiom of choice, one cannot silently identify every infinite input with an aleph, and the asserted conclusion needs those hypotheses.

The Hartogs number $h(X)$ is the least ordinal that does not inject into $X$. It is an initial ordinal: if it were equinumerous with a smaller ordinal $\alpha <h(X)$, composing that bijection with an injection $\alpha \to X$ would contradict its definition. With choice, well-order an infinite $X$ and put $|X|=\kappa ={\aleph }_{\eta }$. Every ordinal below ${\kappa }^{+}$ has cardinal at most $\kappa$ and injects into $X$, whereas ${\kappa }^{+}$ does not. Hence the Hartogs numbers under choice satisfyTo prove regularity, suppose $\mathrm{cf}({\kappa }^{+})=\mu <{\kappa }^{+}$. Since ${\kappa }^{+}$ is a successor cardinal, $\mu \le \kappa$. Choose a cofinal sequence $⟨{\alpha }_i:i<\mu ⟩$ below ${\kappa }^{+}$. Each ${\alpha }_i$ has cardinal at most $\kappa$, and choice givesBut cofinality makes this union all of ${\kappa }^{+}$, a contradiction. Therefore $\mathrm{cf}({\kappa }^{+})={\kappa }^{+}$, and every infinite-input Hartogs value is a regular initial successor aleph. Conversely every successor aleph occurs by taking $X$ to be its predecessor cardinal.