The Hartogs number of $A$ exists for any set $A$.
Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!
My attempt:
Let $H=\{(W,\prec) \in \mathcal{P}(A) \times \mathcal{P}(A\times A) \mid \prec \text{ is a well-ordering on } W\}$. It is well-known that every $(W,\prec)\in H$ is isomorphic to a unique ordinal.
By Axiom of Replacement, $h(A)=\{\alpha \in \text{Ord} \mid (\alpha,<) \text{ is isomorphic to some } (W,\prec) \in H\}$ is a set. It follows that $h(A)$ is a set of ordinals.
Assume $\gamma\in\alpha\in h(A)$. Then there is an isomorphism $f$ between $(\alpha,<)$ and some $(W,\prec) \in H$. Then $f \restriction \gamma$ is an isomorphism between $(\gamma,<)$ and $(f[\alpha],\prec) \in H$. Thus $\gamma \in H$. Hence, $h(A)$ is transitive.
To sum up, $h(A)$ is a set of ordinals and transitive. Thus $h(A)$ is an ordinal. Moreover, $h(A)$ is not equipotent to any subset of $A$. If not, there is a bijecton $g$ from $h(A)$ to some $W \in \mathcal{P}(A)$. We define a well-ordering $\prec$ on $W$ by $w_1 \prec w_2 \iff g^{-1}(w_1) < g^{-1}(w_2)$. As such, $(h(A),<)$ is isomorphic to $(W,\prec)$. Then $h(A)\in h(A)$, which is a contradiction. This completes the proof.