Help on proving equal cardinalities

**gordo151091** · November 18th 2011, 04:30 AM

Hello, I am trying to proof the following claim.
$\left| P(A \times B) \right| = \left| P(B)^A \right|$

Where $P$ is the power set and $P(B)^A := \left \{ \left f: ( f: A \to P(B) \right ) \right \}$ .

I am trying to construct a bijection or two injections and cantors theorem to proof they have the same cardinality, but I seem to be getting nowhere. Can anyone help me?

**Opalg** · November 18th 2011, 04:58 AM

Originally Posted by gordo151091

Hello, I am trying to proof the following claim.
$\left| P(A \times B) \right| = \left| P(B)^A \right|$

Where $P$ is the power set and $P(B)^A := \left \{ \left f: ( f: A \to P(B) \right ) \right \}$ .

I am trying to construct a bijection or two injections and cantors theorem to proof they have the same cardinality, but I seem to be getting nowhere. Can anyone help me?

I would define a bijection $\psi : P(A\times B) \to P(B)^A$ by $\psi(S)(a) = \{b\in B:(a,b)\in S\}$ , for $S\in P(A\times B)$ and $a\in A.$

The inverse map $\phi$ would be given by $\phi(f) = \{(a,b):b\in f(a)\}$ , for $f\in P(B)^A.$

Intuitively, a subset of the product space AxB can be identified with the set of "vertical slices" given by fixed elements of A, as in the picture.

**gordo151091** · November 18th 2011, 07:11 AM

So the $b \in B$ in the red line from picture would be the resulting set by the function $\psi$ for a particular $S$ and $a$

**Opalg** · November 18th 2011, 07:22 AM

Originally Posted by gordo151091

So the $b \in B$ in the red line from picture would be the resulting set by the function $\psi$ for a particular $S$ and $a$

Yes. The blobby area in the picture is supposed to represent a subset S of AxB, and $\psi(S)$ is the function whose value at a point $a$ is the set of second coordinates of all points in S whose first coordinate is $a.$

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