This was inspired by this recent [question](https://mathoverflow.net/questions/46970/proofs-of-the-uncountability-of-the-reals). 


In my [answer](https://mathoverflow.net/questions/46970/proofs-of-the-uncountability-of-the-reals/47022#47022) there, I pointed out that, given $F:{\mathcal P}(X)\to X$, an argument dating back to Zermelo allows us to define a pair $(A,B)$ of distinct subsets of $X$ witnessing that $F$ is not injective. The pair is defined in terms of a well-ordering that is constructed using $F$.


Of course, the usual Cantor argument also shows that $F$ is not injective: One considers $$ A=\{F(Z)\mid F(Z)\notin Z\},$$ and argues that there must be a $B$ with $A\ne B$ and $F(A)=F(B)$.


My question is:



> 
> Can we exhibit such a set $B$ (definably from $F$)?
> 
> 
> 



Cantor's argument revisited
If I understood the OP correctly, the problem can be stated as follows :


**Problem 1.** Let $X$ be a set, let $F:{\cal P}(X) \to X$, and let $A$ be defined
as above: $$A=\lbrace F(Z) | Z\subseteq X, F(Z) \not\in Z\rbrace.$$ Find a definable $B$ (in terms of $F$) such that $B \neq A$ and $F(B)=F(A)$.


Now Problem 1 is equivalent to the simpler problem :


**Problem 2.** Let $Y$ be a set and let $Y'$ be a nonempty subset of $Y$. Find
a definable $y\_0$ (in terms of $Y$ and $Y'$) which is in $Y'$.


The interesting and nontrivial part of the equivalence is of course, to show that we can solve Problem 2 if we can solve Problem 1. Here is how. Let $Y$ and $Y'$ be as above. Take two elements $a,b$ and a countable set $W=\lbrace w\_k \rbrace\_{k \geq 0}$ outside of $Y$. Now define
$X$ to be the disjoint union of $\lbrace a,b \rbrace$ and $Y \times W$, and define
$F : {\cal P}(X) \to X$ by:


1. $F(\lbrace (y,w\_0) \rbrace)=a$, if $y\in {Y'} $,
2. $F(\lbrace (y,w\_{k+1}) \rbrace)=(y,w\_k)$ , for all $y\in Y$ and $k\ge0$,
3. $F(X)=a$, and
4. $F(Z)=b$ for all other subsets $Z$ of $X$ (thus $F(\emptyset)=b$).


Now, by construction, $A=X$, and any solution $B$ to Problem 1 is of the form $\lbrace (y,w\_0) \rbrace$ for some $y\in Y'$, thereby solving Problem 2.


