I have to prove that if the cardinality of A = the cardinality of B AND the cardinality of C = the cardinality of D then the cardinality of (A x C) = the cardinality of (B x D) with sets A,B,C,D and x denoting the cartesian product.
I know that there exists a bijection f: A to B and a bijection g: C to D. But how do I proceed using this idea of bijections?
Thanks for your help in advance.
Can I say this?
I know that there is a function f: A to B that is injective, a function g: B to A that is injective, a function h: C to D that is injective, and a function j: D to C that is injective (since the cardinality of X <= the cardinality of Y if there exists an injective function from X to Y).
Define the function k: (A x C) to (B x D).
Do the statements above imply that a function l: (A x C) to (B x D) is injective with l(a,c) = (a,c) and a function m: (B x D) to (A x C) is injective with m(b,d) = (b,d). Then by the Cantor-Bernstein Theorem, the cardinality of (A x C) is equal to the cardinality of (B x D).
PvtBillPilgrim, did you read the first reply by TPH?
Why would you go to the lengths you have tried? I cannot follow them.
Look, we are given that A & B have the same cardinality so the is a bijection between them. Same statement for C & D. As was pointed out by THP it is simple to show that is a bijection.
Well I don't understand exactly what that means.
Is that the proof or do I need to show that the function h in his post is a bijection? I understand if it's a bijection, then the cardinality is the same, but I don't see why it's a bijection. Is it just obvious?
Basically, how do I prove that the function h defined by h(a,b) = (f(a),g(b)) is a bijection?
If h(a,b)=h(a',b'), it means that (f(a),g(b))=(f(a'),g(b')) --> f(a)=f(a') and g(b)=g(b'). Since f and g are bijections, a=a' and b=b'. Thus h is an injection.
Let y in . To prove it's a surjection, you have to prove that there exist a and b in such that h(a,b)=y.