Hartogs bounds in iterated power sets

For each injection $f:\alpha \to X$, let $T_f=\{f[\gamma ]:\gamma \le \alpha \}\subseteq \mathcal{P}(X)$. Inclusion well-orders this chain in type $\alpha +1$, because the initial images strictly increase. For each $\alpha <h(X)$, let $B_{\alpha }$ be the set of all such chains of type $\alpha +1$ arising from these injections. It is a nonempty element of ${\mathcal{P}}^3(X)$, and distinct indices give disjoint such sets because a chain has unique order type. Thus $\alpha \mapsto B_{\alpha }$ is injective. For $\alpha =0$ the chain is $\{\mathrm{\varnothing }\}$.

Let $R_{\alpha }$ be the set of all reflexive well-order relations on subsets of $X$ of type $\alpha$. Reflexivity makes the underlying set recoverable from the diagonal, including singleton orders; the empty order has the empty relation. These are nonempty pairwise disjoint subsets of $\mathcal{P}(X^2)$. Thus $h(X)⪯{\mathcal{P}}^2(X^2)$, and a supplied bijection $X^2\approx X$ gives the double-power bound.

If $X$ is infinite and Dedekind-finite, then $h(X)=\omega$: every finite ordinal embeds by finite induction, while omega does not. Each $[X{]}^n$ is nonempty, and these sets of subsets are disjoint for distinct $n$. Hence $n\mapsto [X{]}^n$ injects omega into ${\mathcal{P}}^2(X)$. If $X$ has finite size $n$, $h(X)=n+1$ and $n+1\le 2^{2^n}$ by elementary finite induction; choose an enumeration of this single finite set to realize the injection. This includes $n=0$.

The first bound implies $h(X)<h({\mathcal{P}}^3(X))$ by the definition of Hartogs. For the other strict bound, singleton inclusion gives $h(X)⪯{\mathcal{P}}^4(X)$. Equality would give an injection $\mathcal{P}(A)\to A$ with $A={\mathcal{P}}^3(X)$. Its inverse on its range, extended elsewhere by $\mathrm{\varnothing }$, would be a surjection $s:A\to \mathcal{P}(A)$, impossible since $\{a\in A:a\mathrm{\notin }s(a)\}$ is missed. Thus the comparison is strict.