Question

Prove:

Answer 1

Let A, B, C, D be sets such that:
if
\[A \approx C\]and \[B \approx D\] then \[AXB \approx CXD\]

Answer 2

@zzr0ck3r

Answer 3

A X B is a cartesian product..

Answer 4

@zzr0ck3r how can I prove that there is a bijective function h: A X B -> x X D?

please help me on showing that they're both injective and surjective...

Answer 5

\[A\approx C\\\text{so there exists a one to one and onto function from A to C}\\\text{we will call it }f(A)\\B\approx D\\\text{so there exists a one to one and onto function from B to D}\\\text{we will call it }g(B)\\\text{define } h:(A\times B)\rightarrow(C \times D),\space h(a,b)=(f(a),g(b))\\\text{to show that h is onto and one to one}\\\text{assume }h(a_0,b_0)=h(a_1,b_1)\\\text{then by definition}\\(f(a_0),g(b_0))=(f(a_1),g(b_1))\\\text{since }f,g\space \text{ are both one to one, we have that }\\(a_0,b_0)=(a_1,b_1)\\\text{thus h is one to one}\\\text{let }(c,d)\in\space C\times D\\\text{since f and g are onto we have that }(c,d)=(f(a),g(b))\text{ for some } a,b \in A, B\\\text{thus }h\text{ is onto and one to one and }A\times B \approx \space C\times D\]