power set question

• Apr 17th 2010, 07:53 AM
jvignacio
power set question
Hey guys i have, A = {1,2,3,4} and B = {2,3,5}, list the following elements:

1. Powerset(A) \ Powerset(B)

Does that mean, find Powerset(A) first then Powerset(B) then if Powerset(A) has subsets the same as Powerser(B), then remove those sets. Then the remaining subsets of Powerset(A) will = Powerset(A) \ Powerset(B) ?

thanks for the help!
• Apr 17th 2010, 08:04 AM
Plato
If $\displaystyle C\in \left(\mathcal{P}(A)\cap \mathcal{P}(B)\right)$ then remove $\displaystyle C$ from $\displaystyle \mathcal{P}(A).$
• Apr 17th 2010, 08:07 AM
jvignacio
Quote:

Originally Posted by Plato
If $\displaystyle C\in \left(\mathcal{P}(A)\cap \mathcal{P}(B)\right)$ then remove $\displaystyle C$ from $\displaystyle \mathcal{P}(A).$

i dont think $\displaystyle \mathcal{P}(A) \cap \mathcal{P}(B)$ is the same as $\displaystyle \mathcal{P}(A) \backslash \mathcal{P}(B)$?
• Apr 17th 2010, 08:07 AM
MoeBlee
Quote:

Originally Posted by jvignacio
if Powerset(A) has subsets the same as Powerser(B), then remove those sets

Maybe you meant, if P(A) has MEMBERS in common with P(B) then remove those common members from P(A).

S\T = {x | x in S and x not in T}

So:

P(A)\P(B)
=
{x | x in P(A) and x not in P(B)}
=
{x | x subset of A and x not subset of B}

I.e., gather the subsets of A to form P(A), then gather the subsets of B to form P(B), then remove those members of P(A) that are also members of P(B) to form P(A)\P(B).
• Apr 17th 2010, 08:10 AM
jvignacio
Quote:

Originally Posted by MoeBlee
So gather the subsets of A, then gather the subsets of B, then from the gathering of subsets of A remove those that are also subsets of B.

that's what I meant :) remove the subsets of A that are also subsets of B
• Apr 17th 2010, 08:13 AM
Failure
Quote:

Originally Posted by jvignacio
i dont think $\displaystyle \mathcal{P}(A) \cap \mathcal{P}(B)$ is the same as $\displaystyle \mathcal{P}(A) \backslash \mathcal{P}(B)$?

That wasn't what Plato was claiming. Imho, he was just making the suggestion that because of

$\displaystyle \mathcal{P}(A)\backslash \mathcal{P}(B)=\mathcal{P}(A)\backslash \Big(\mathcal{P}(A)\cap \mathcal{P}(B)\Big)$

it would by easier to determine the solution of your exercise by using the right side of this identity.
• Apr 17th 2010, 08:15 AM
jvignacio
Quote:

Originally Posted by Failure
That wasn't what Plato was claiming. Imho, he was just making the suggestion that because of

$\displaystyle \mathcal{P}(A)\backslash \mathcal{P}(B)=\mathcal{P}(A)\backslash \Big(\mathcal{P}(A)\cap \mathcal{P}(B)\Big)$

it would by easier to determine the solution of your exercise by using the right side of this identity.

Yep I understand now. thanks!
• Apr 17th 2010, 08:18 AM
MoeBlee
Quote:

Originally Posted by jvignacio
i dont think $\displaystyle \mathcal{P}(A) \cap \mathcal{P}(B)$ is the same as $\displaystyle \mathcal{P}(A) \backslash \mathcal{P}(B)$?

Correct. But what the poster Plato said is correct.

Let 'n' stand for the binary intersection operation.

S\T = S\(SnT)

P(A)\P(B) = P(A)\(P(A)nP(B))

/

Edit: I see that you got it now anyway.
• Apr 17th 2010, 08:42 AM
jvignacio
Quote:

Originally Posted by MoeBlee
Correct. But what the poster Plato said is correct.

Let 'n' stand for the binary intersection operation.

S\T = S\(SnT)

P(A)\P(B) = P(A)\(P(A)nP(B))

/

Edit: I see that you got it now anyway.

In a case where it was: find elements in: $\displaystyle A^2 \cap B^2$, would it be ZERO elements? since the set A deals with 4 numbers and set B deal with 3, therefore never having any subsets in common. ?
• Apr 17th 2010, 12:52 PM
Swlabr
Quote:

Originally Posted by Failure
That wasn't what Plato was claiming. Imho, he was just making the suggestion that because of

$\displaystyle \mathcal{P}(A)\backslash \mathcal{P}(B)=\mathcal{P}(A)\backslash \Big(\mathcal{P}(A)\cap \mathcal{P}(B)\Big)$

it would by easier to determine the solution of your exercise by using the right side of this identity.

I think it was more that as $\displaystyle \mathcal{P}(A)\not\subset \mathcal{P}(B)$ you can't really talk about $\displaystyle \mathcal{P}(A)\backslash \mathcal{P}(B)$, and instead we should talk about $\displaystyle \mathcal{P}(A)\backslash \Big(\mathcal{P}(A)\cap \mathcal{P}(B)\Big)$. For notation reasons, however, we use the former as it is "neater".
• Apr 17th 2010, 02:14 PM
Plato
Quote:

Originally Posted by Swlabr
I think it was more that as $\displaystyle \mathcal{P}(A)\not\subset \mathcal{P}(B)$ you can't really talk about $\displaystyle \color{red}\mathcal{P}(A)\backslash \mathcal{P}(B)$

Why not? Set difference is set difference.
To say that $\displaystyle C\in\mathcal{P}(A)\backslash \mathcal{P}(B)$ means $\displaystyle C\in\mathcal{P}(A) \text{ and }C\notin \mathcal{P}(B)$.
Moreover $\displaystyle \mathcal{P}(A)\not\subset \mathcal{P}(B)$ has nothing to do with the general concept.