Let $A$ be a set and let $\mathcal{H}(A)$ be the Hartogs number of $A$. Show that $|\mathcal{H}(A)|\neq|\mathcal{P}(\mathcal{P}(A\times A))$.
Proof attempt: By contradiction. Suppose that $|\mathcal{H}(A)|=|\mathcal{P}(\mathcal{P}(A\times A))|$. This would mean there is a bijective function $f:\mathcal{P}(\mathcal{P}(A\times A))\longrightarrow\mathcal{H}(A)$. Now, here's where I'm getting stuck. My first thought is that if there was a bijection from $\mathcal{P}(\mathcal{P}(A\times A))\longrightarrow\mathcal{H}(A)$, then that would mean that every set of relations on $A$ would be well-orderable. Is that the case? If not, please tell me where I'm going wrong and maybe give me a hint on what to try next?
Thanks in advance for your help.
HINT: Suppose that $\mathscr{H}(A)=|\wp(\wp(A\times A))|$; then there is a bijection $$h:\mathscr{H}(A)\to\wp(\wp(A\times A))\;.$$ You already know that there is a surjection $$f:\wp(A\times A)\to\mathscr{H}(A)\;.$$ You would then have a surjection
$$h\circ f:\wp(A\times A)\to\wp(\wp(A\times A))\;.$$