Bijection with empty intersection Let $A$ be an arbitrary set. How can we construct a set $B$, in bijection with $A$, such that 
$A \cap B=\emptyset$?
 A: While the answers suggesting $A\times\{A\}$ are correct, they do rely on the axiom of regularity, and this problem can be solved without it.
Let $x$ be a set such that there is no ordered pair $\langle x,y\rangle\in A$ (for any $y$, that is). Why does this $x$ exist? Because the projection of $A$ onto the right coordinate (of all its elements which are ordered pairs) is a set; therefore it does not include all sets.
Now take $B=\{x\}\times A$. Clearly every element in $B$ is an ordered pair with $x$ in the right coordinate, so $B\cap A=\varnothing$.

Of course, assuming the axiom of regularity holds we can prove that $x=A$ is a valid choice, so it is easier in that case.
A: Is the required set arbitrary? If you work in the standard model (well-founded), you can make the set
$$
B = \{ \{x, A\} : x\in A \}
$$
no element y of B is an element of A, otherwise we would have $A\in y\in A$, and there is a natural bijection between $A$ and $B$ given by
$$
f(x) = \{x, A\}
$$
A: Define $B=A\times\{A\}$ and $f:A\rightarrow B$ by $a\mapsto\left(a,A\right)$.
$\left(a,A\right)=\left\{ \left\{ a\right\} ,\left\{ a,A\right\} \right\} \in A$
leads to $A\in\left\{ a,A\right\} \in A$ wich cannot be true (if the
axiom of regularity is accepted). So $A\cap B=\emptyset$.
