power set question

**jvignacio** · April 17th 2010, 07:53 AM

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!

**Plato** · April 17th 2010, 08:04 AM

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

**jvignacio** · April 17th 2010, 08:07 AM

Originally Posted by Plato

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

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

**MoeBlee** · April 17th 2010, 08:07 AM

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).

**jvignacio** · April 17th 2010, 08:10 AM

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

**Failure** · April 17th 2010, 08:13 AM

Originally Posted by jvignacio

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

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

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

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

**jvignacio** · April 17th 2010, 08:15 AM

Originally Posted by Failure

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

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

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

Yep I understand now. thanks!

**MoeBlee** · April 17th 2010, 08:18 AM

Originally Posted by jvignacio

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

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.

**jvignacio** · April 17th 2010, 08:42 AM

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: $A^2 \cap B^2$ , 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. ?

**Swlabr** · April 17th 2010, 12:52 PM

Originally Posted by Failure

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

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

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 $\mathcal{P}(A)\not\subset \mathcal{P}(B)$ you can't really talk about $\mathcal{P}(A)\backslash \mathcal{P}(B)$ , and instead we should talk about $\mathcal{P}(A)\backslash \Big(\mathcal{P}(A)\cap \mathcal{P}(B)\Big)$ . For notation reasons, however, we use the former as it is "neater".

**Plato** · April 17th 2010, 02:14 PM

Originally Posted by Swlabr

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

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

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