# Bijection with empty intersection [closed]

Let $A$ be an arbitrary set. How can we construct a set $B$, in bijection with $A$, such that $A \cap B=\emptyset$?

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## closed as off-topic by Claude Leibovici, AlexR, egreg, Michael Grant, Eric StuckyMar 21 '14 at 13:03

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Please share your thoughts so far :) – Shaun Mar 21 '14 at 11:03

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.

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This is exactly why I didn't like the other answers, but I had to go away, so unfortunately I didn't have time to post my answer or explain anything. – user2345215 Mar 21 '14 at 12:17
No biggie. I got your message and came to post this for you. ;-) – Asaf Karagila Mar 21 '14 at 12:24
As always, the most complete response is from @AsafKaragila :) – rewritten Mar 21 '14 at 13:16
@rewritten: Hehe, I don't know about that, I leave a lot of incomplete hints over this website. But thanks! :-) – Asaf Karagila Mar 21 '14 at 13:20

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 $$f(x) = \{x, A\}$$

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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 $A\cap B=\emptyset$.

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If you downvote then tell me why. I upvoted the answers of Git Gud and rewritten because they are okay and also to compensate. – drhab Mar 21 '14 at 11:03
Thank you for suppressing my opinion and voting based on other people's votes. – user2345215 Mar 21 '14 at 12:20