# Homework Help: 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

http://suprfile.com/src/1/3m7bfrq/lastscan.jpg [Broken]

Last edited by a moderator: May 2, 2017
2. Oct 7, 2006

### Hurkyl

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

3. Oct 7, 2006

### HallsofIvy

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

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!