BxB, sets question, need to verify if this is right.

1. Oct 7, 2006

mr_coffee

Hello everyone, i was wondering if this is correct or not.
I circled the 2 I was iffy about. Thanks!
Note: P stands for power sets, and X stands for Cartesian Product

2. Oct 7, 2006

Hurkyl

Staff Emeritus
The two you circled are, in fact, wrong.

3. Oct 7, 2006

HallsofIvy

Staff Emeritus
Remember that AxA, AxB, BxB, etc. are ordered sets.

4. Oct 7, 2006

Hurkyl

Staff Emeritus

5. Oct 8, 2006

mr_coffee

I went over the chapter again and I still can't see what I did wrong, especially in the above one where i'm finding the power set. Can you give me some feedback on why its wrong or how I can correct it?

For the BxB would it just be B? {a,b}?

6. Oct 8, 2006

0rthodontist

You want to take the power set of
$$\{\phi, \{\phi\}\}$$
It may be a little less confusing if you instead try to find the power set of {x, y} where x = $$\phi$$ and y = $$\{\phi\}$$, then substitute back.
in the second one you want to find {a, b} x {a, b}. Can you find {a, b} x {c, d} and then let c = a, d = b?
In these one thing to check is the size of the set in your answer. If A and B are finite, |P(A)| = 2|A| and |A x B| = |A| x |B|

Last edited: Oct 8, 2006
7. Oct 8, 2006

HallsofIvy

Staff Emeritus
What a remarkably stupid thing for me to say! It must have been past my bedtime. AxA, AxB, BxB, etc. are sets of ordered pairs. In other words, if a, b are both in B, BxB will contain both (a,b) and (b,a).

8. Oct 8, 2006

mr_coffee

Thanks for the responces...
In other words, if a, b are both in B, BxB will contain both (a,b) and (b,a).
So it would just be BxB = { (a,b), (b,a) } ?
Why wouldn't it include any of the others? like
BxB = {(a,a),(a,b),(b,a),(b,b) }
If BXB is an ordered pair that means AXB is, and when you do AXB in this example you do alot more grouping like for instance:

A = {x,y,z,w}
B = {a,b}
AxB = { (x,a), (y,a),(z,a),(w,a),(x,b),(y,b),(z,b),(w,b) }
You group up each A element with a B.
So why wouldn't u also do that with BxB? group each B element with itself?
(a,a),(b,b),(a,b),(b,a)

Ortho with ur hint,
to let c = a, d = b. i get the following:
{a,b} x {c, d} = { (a, c), (a, d), (b, c), (b, d) }
now sub back in
BxB = { (a, a), (a, b), (b, a), (b, b) }

Last edited: Oct 8, 2006
9. Oct 8, 2006

mr_coffee

Also for the power sets i took ur advice ortho and got the following:

P({null,{null}})
let x = null
let y = {null}
P({x,y}) = {null,{x},{y},{x,y}}
now sub in for x and y and i got:
P({x,y}) = {null,{null}, { {null} }, { null,{null} } }

10. Oct 8, 2006

0rthodontist

Those are both right.

11. Oct 8, 2006

mr_coffee

o yeah! thanks for the help again!