Hi

here's a problem I am trying to solve. Suppose A,B,C,D are sets

$\displaystyle [(A\sim B)\wedge(C\sim D)\Rightarrow (^AC \sim \; ^BD)]$

where $\displaystyle ^AC$ is set of all functions from A to C. and $\displaystyle ^BD$ is

set of all functions from B to D. Following is my work. I have already prove that

$\displaystyle [((A\sim B)\wedge(C\sim D)\Rightarrow(A\times C \sim B \times D)] $

Suppose

$\displaystyle (A\sim B)\wedge(C\sim D)$

$\displaystyle \therefore A\times C \sim B \times D $

I have also already proved that, for any sets A and B,

$\displaystyle (A\sim B)\Rightarrow(\mathcal{P}(A)\sim \mathcal{P}(B))$

$\displaystyle \therefore \mathcal{P}(A\times C) \sim \mathcal{P}(B \times D)$

that means that there is a bijection from $\displaystyle \mathcal{P}(A\times C)$

to $\displaystyle \mathcal{P}(B \times D)$.

$\displaystyle \mathcal{P}(A\times C)$ is the set of all subsets of $\displaystyle A\times C$

. But subsets of $\displaystyle A\times C$ are just relations from A to C. So

$\displaystyle \mathcal{P}(A\times C)$ is set of all relations from A to C. In similar way,

$\displaystyle \mathcal{P}(B\times D)$ is set of all relations from B to D. Functions are special kind of relations , So

$\displaystyle ^AC\subseteq \mathcal{P}(A\times C)$

$\displaystyle ^BD \subseteq \mathcal{P}(B\times D)$

Since $\displaystyle \mathcal{P}(A\times C) \sim \mathcal{P}(B\times D)$ , there is some bijection between these two sets. Let it be h

$\displaystyle h:\mathcal{P}(A\times C)\to \mathcal{P}(B\times D)$

From here on, I am stuck. We have a bijection between one family of relations to another family of relations. What we need is a bijection between the subsets of

these families.