# Bijection between AxB and BxA

1. Dec 27, 2008

### Unassuming

1. The problem statement, all variables and given/known data

Show that there is a bijective correspondence of AxB with BxA.

2. Relevant equations

3. The attempt at a solution

I am lacking the general understanding of this. Can I create a function g such that,

g: (a in A, b in B) --> (b in B, a in A).

If A and B are sets, then are AxB and BxA the same?

2. Dec 27, 2008

### Dick

You just did create such a function. If (a,b) is in AxB, then let g((a,b))=(b,a), which is what you said, if not what you meant. (b,a) is in BxA. Now you just have to prove it's a bijection

3. Dec 27, 2008

### HallsofIvy

Staff Emeritus
Yes, you can do that. Now you have to prove that is a "bijection": its both "one-to-one" (an injection) and "onto", (a surjection).

No, absolutely not! But there exist an injection between them so they have the same "cardinality". For example, if A= {x} and B= {y} then A x B= {(a, b)} and B x A= {(b, a)}. Those two sets have the same cardinality (1) but are different sets because they contain different pairs: as an ordered pair, (a, b) is NOT the same as (b, a).

4. Dec 28, 2008

### knollen baum

But only if WLOG A is not an empty set, because the Zermelo–Fraenkel set theory (the standard set theory) defines in the "axiom of empty set" that there isn't an a in A.

$$\exists X\, \forall y\, \lnot (y \in X)$$

And do not forget to make your answer save to sets with infinite cardinality.

5. Dec 28, 2008

### Hurkyl

Staff Emeritus
It's not clear exactly what you're saying. But it doesn't seem to be relevant to anything anyone else said.

6. Dec 28, 2008

### knollen baum

Sorry, my english is not very good, but i will try to explain it.

You can not prove a bijective correspondence of AxB with BxA for any set A, B by an function that says nothing for empty sets.

The function says:
g: (a in A, b in B) --> (b in B, a in A)

The "axiom of empty set" says:
$$\exists X\, \forall y\, \lnot (y \in X)$$
It means that there is an set A with no a in A. We call it emtpy set.

If you look again at the function g you will see that it is not defined for empty sets. Because if WLOG (Without loss of generality) A is the empty set, there is no a in A and so you dont know what (a in A, b in B) --> (b in B, a in A) means, or do you know what (,) -> (,) means?

So if you prove that g: (a in A, b in B) --> (b in B, a in A) is bijective, than you only know that there is a bijective correspondence of AxB with BxA when neither A nor B is an empty set.
Sure it is trivial that there is a bijective correspondence of {} and {}, but if you have to answer an basic question it should be an complete anwser without the "trivial"-option.

If i have to correct such an answer i would mark this point red, so i wanted to show this point before anyone will really mark this answer red.

Last edited: Dec 28, 2008
7. Dec 28, 2008

### Citan Uzuki

Actually, the function definition $f:A \times B \rightarrow B \times A: (a,\ b) \mapsto (b,\ a)$ is perfectly valid even in the case where at least one of A or B are empty. Why? Because in that case, $A \times B = \emptyset$, so it is vacuously true that every element in $A \times B$ is of the form (a, b) for some $a \in A$ and some $b \in B$. Thus, there is no need to handle the case where one of A or B are empty separately, since the general argument applies to it as well (and in that case, the bijection constructed is simply the empty function).

8. Dec 29, 2008

### Hurkyl

Staff Emeritus
You're use of 'WLOG' here is incorrect; assuming A to be empty is a loss of generality.

9. Dec 29, 2008

### HallsofIvy

Staff Emeritus