Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Claude Leibovici, AlexR, egreg, Eric Stucky
If this question can be reworded to fit the rules in the help center, please edit the question.

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.

share|cite|improve this answer
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 $$ f(x) = \{x, A\} $$

share|cite|improve this answer

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 $A\cap B=\emptyset$.

share|cite|improve this answer
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

Not the answer you're looking for? Browse other questions tagged or ask your own question.