For every set $A$, there exists a well ordered set $V$ such that there exists no surjection $\pi: A \rightarrow V$. I am proving that for every set$ A$, there exists a well ordered set $X$ such that there exists no surjection $\pi : A \longrightarrow V$.
I think that's very simple, and so I think maybe I made some mistake. Here it is what I try:
By Cantor theorem I know that there exists no surjection between $A$ and $\mathcal{P}(A)$, and, as for every set $X$, there exists a well order with domain $X$. There is no surjection between $A$ and the well ordered set $(\mathcal{P}(A),\leq_{\mathcal{P}(A)})$
Am I right?
Please correct me. Thank you.
 A: What you wrote is completely indecipherable. You use $X$ twice, and it's not clear whether or not you make an appeal to the axiom of choice (or the well-ordering principle), which is something that you really shouldn't do.
This is very similar to Hartogs theorem, only with surjections, and you can repeat the essence of the proof of Hartogs theorem, using prewellorderings of subsets of $A$ rather than well-orderings of subsets of $A$.
But a simpler proof would use Hartogs theorem, in the following way:
Note that if $A$ can be surjected onto $\alpha$, then $\alpha$ can be injected into $\mathcal P(A)$. So if $A$ can be mapped onto every ordinal, then every ordinal can be injectively mapped into $\mathcal P(A)$. But Hartogs theorem tells us that there is an ordinal $\alpha$ which cannot be injectively mapped into $\mathcal P(A)$, and therefore there is an ordinal that $A$ cannot be mapped surjectively onto. $\square$
(The theorem itself is due to Lindenbaum, but it first appeared in Sierpinski's book from 1948.)
A: As you've noted, there is no surjection between $A$ and $\mathcal{P}(A)$. It suffices to show that $\mathcal{P}(A)$ can be well-ordered. But the Axiom of Choice is equivalent to the Well-Ordering Theorem, which states that every set can be well-ordered. In particular, $\mathcal{P}(A)$ can be well-ordered, so we are done.
