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I think that the result is the empty set.

\(\not \exists x\in S\;|\; x\in\wp S \)

Am I right?

@brinethery what do you think?

If \[\wp S\] is the power set, then list out everything in the power set but kick out the set S

That is what I did.

ok, but you won't get the empty set

But I didn't kick out the set S.

that's the what the " - S " portion is saying

What I think it says is that I must kick out the elements of S that are elements of P(S)

and none of the elements of S are in P(S)

\(1\not \in S\) for example

because \(1\not = \{1\}\)

And the same happens with 2,3,4,5

Mmm then I think \(\wp S - S = \wp S\)

not quite, P(S) - S does not equal P(S)

close though, because you're only erasing one element of the powerset

I cant quit {1,2,3,4,5} because \(\{1,2,3,4,5\}\not \in S\)

{1,2,3,4,5} is S, so it's a subset of P(S)
S is a subset of P(S)

heres a page on the power set
http://en.wikipedia.org/wiki/Power_set

I'm agree with the power set expansion. But I think you're difference is wrong

P(S) - S means "start with P(S), then remove the set S itself since S is an element of P(S)"

\(B-A=\{x\in B | x\not \in A\}\)

What are you saying is this: \(\wp S - \{S\} \)

oh good point, I'm mixing up sets and elements here...

so P(S) - S would be start with P(S) and kick out any elements you find in S

Yeah, and there is no elements that are in both sets.

so then your original answer of empty set works

you can't quit \(S\) because \(S\not \in S\)

Yeah but why are you removing S?

When you do a set difference you remove the elements that both sets have in common right?

So you're removing nothing from \(\wp S\)

You can't remove 1 because this element is not in \(\wp S\)

\(1\not = \{1\}\) for example

true, not thinking on this one lol

lol no problem.

So do you agree with my answer @jim_thompson5910 ?\[\wp S - S = \wp S\]

yes i do

there's no element in S that's in P(S), so you're taking out nothing from P(S) when you do P(S) - S

What if A = {1, {1}}, what would its power set look like?

probably this, but I'm not 100% sure
P(S) =
{
{1,{1}}
{1}, {{1}}
{}
}

\[\wp A = \{\emptyset, \{1\}, \{\{1\}\}, A\}\]

In that instance, the original set contains the element {1}, but so does the power set.

Yes you're right @LogicalApple that relation holds for sets with non-set elements.

I mean \(\wp A - A \not = \wp A\)

I was thinking along the same lines

You can take out the set itself if you have the following expression:
\[\wp S - \{S\}\]

This problem is tricky

Ohhh, I read it wrong then.

Except perhaps the empty set?

yeah exactly !

not, the empty set is not in S

Yeah, in this case it there isn't any elements in common, but I was just thinking generally.

mmm ok.

I'm used to seeing the power set denoted with \mathcal P : \(\mathcal P(S) \)

Yeah there are several way of writting it. I just use the same as the one in my book hehe.

Not at all, but that's what the question wants.

Because \(S\not \in S\)

I agree with \(S\subseteq\wp S\).

Ok. now list all the elements of S.

And quit the elements that are in S and the power set of S

Note that P(S) does not have any element like 1,2,3,... , only subsets.

brb

ok

So P(S)\S will be left with only 31 subsets, the first 31.

* 31 elements.

Why you're removing the set S?

S is not an element of S, so you can't remove it.

You must remove from the power set of S the elements of S that are in the power set of S too.

yeas. the power set of S has set elements. then you can't remove non set element from it.

Is 1 a subset of {1, 2} ?

1 is not a set. so it can't be a subset.

yeah you're right @LogicalApple

In fact, in general, P(S)-S has 2^n-1 elements.

Remember the definition of a set difference: \[A-B=\{x\in A|x\not \in B\}\]

I'm not looking for the number of elements. but for a resulting set.

brb

empty set belongs to both P(S) and S, but not 1,2,3...5, so these don't need to be excluded.

The power set was generated by wolframalpha

Note that 1 is an element of S, but {1} is not, it is a subset of S.

*P also has {}, but S does not.

Great job @LogicalApple Thank you!

I don't understand what the problem here is...
\[
\mathcal P(S) - S = \mathcal P(s)
\]

that expression hold for \(S = \{1,2,3,4,5\}\)

Yeah.

I would think that expression holds unless S contains an element that is also a subset of S.

It holds when: \[
\neg \exists A \subseteq S \wedge A \in S
\]
Why is this question not closed?

It is actually closed haha.

I mean, why so many responses still?

{} is a subset of S, that's how it got into P(S)!
In fact, {} is a subset of every set.

Therefore, by definition of \ , {} gets kicked out of P(S)-S.

The empty set can't be removed because is not an element of S neither S.

Look a this its more clear.

For example \( \emptyset \subset S\) but \(\emptyset \not \in S \)

Then it's not correct to remove the empty set from the powerset in this case right?

No the empty set is not in every set. The empty set is a subset of every set.