- #1

cgjolberg

- 2

- 0

## Homework Statement

For every pair of sets (A,B) we have P(AxB)=P(A)xP(B)

Prove or disprove the above statement.

## Homework Equations

## The Attempt at a Solution

I have attempted solving this using A={1,2} and B={a,b}

AxB={(1,a),(1,b),(2,a),(2,b)}

P(AxB)={ø,{(1,a)},{(1,b)},{(2,a)},{(2,b)},{(1,a),(1,b)},{(1,a),(2,a)},{(1,a),(2,b)},{(1,b),(2,a)},{(1,b),(2,b)},{(2,a),(2,b)},{(1,a),(1,b),(2,a)},{(1,a),(1,b),(2,b)},{(1,a),(2,a),(2,b)},{(1,b),(2,a),(2,b)},{(1,a),(1,b),(2,a),(2,b)}}

P(A)={ø,{1},{2},{1,2}}

P(B)={ø,{a},{b},{a,b}}

P(A)xP(B)={(ø,ø),(ø,{a}),(ø,{b}),(ø,{a,b}),({1},ø),({1},{a}),({1},{b}),({1},{a,b}),({2},ø),({2},{a}),({2},{b}),({2},{a,b}),({1,2},ø),({1,2},{a}),({1,2},{b}),({1,2},{a,b})}

So from this it seems clear that they are not equal. The easiest way to state that without all of the work I just did seems to be that P(AxB) creates sets of sets, whearas P(A)xP(B) creates pairs of sets or something along those lines.

However, from several websites I have found through searching, several say these two statements are equivalent, so now I am confused.

Also, am I correct in crossing the ø through like I did in P(A)xP(B)? I am not sure if you are suppose to do it like I did or just have one ø at the beginning of the answer.

Thanks for the help!