Question

Given the following set \(S:=\{1,2,3,4,5\}\) evaluate the following expression:
\[\wp S - S\]

Answer 1

I think that the result is the empty set.

Answer 2

\[\wp S - S = \emptyset\] because there is not an element on S that equals an elemento of the power set of S.

Answer 3

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

Answer 4

Am I right?

Answer 5

@brinethery what do you think?

Answer 6

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

Answer 7

That is what I did.

Answer 8

ok, but you won't get the empty set

Answer 9

But I didn't kick out the set S.

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

If P(S) is the powerset of S, then

P(S) = { list of all subsets of S}

P(S) =
{
{1,2,3,4,5}
{1,2,3,4}, {1,2,3,5}, {1,2,4,5}, {1,3,4,5}, {2,3,4,5}
{1,2,3}, {1,2,4}, {1,2,5}, {1,3,4}, {1,3,5}, {1,4,5}, {2,3,4}, {2,3,5}, {2,4,5}, {3,4,5}
{1,2}, {1,3}, {1,4}, {1,5}, {2,3}, {2,4}, {2,5}, {3,4}, {3,5}, {4,5}
{1}, {2}, {3}, {4}, {5}
{}
}

Answer 14

you then kick out S itself to get this

P(S) - S =
{
{1,2,3,4}, {1,2,3,5}, {1,2,4,5}, {1,3,4,5}, {2,3,4,5}
{1,2,3}, {1,2,4}, {1,2,5}, {1,3,4}, {1,3,5}, {1,4,5}, {2,3,4}, {2,3,5}, {2,4,5}, {3,4,5}
{1,2}, {1,3}, {1,4}, {1,5}, {2,3}, {2,4}, {2,5}, {3,4}, {3,5}, {4,5}
{1}, {2}, {3}, {4}, {5}
{}
}

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

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

S is a subset of P(S)

Answer 23

P(S) = { all possible subsets of S }

P(S) =
{
S,
{sets of size n-1 where each element is in S}
{sets of size n-2 where each element is in S}
.....
.....
{sets of size 2 where each element is in S}
{sets of size 1 where each element is in S}
{set of size 0...ie the empty set}
}

Answer 24

heres a page on the power set

http://en.wikipedia.org/wiki/Power_set

Answer 25

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

Answer 26

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

Answer 27

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

Answer 28

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

Answer 29

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

Answer 30

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

Answer 31

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

Answer 32

so then your original answer of empty set works

Answer 33

No, I think the right answer is \(\wp S - S = \wp S\) because there is no elements to quit from \(\wp S\)

Answer 34

We have to note that the elements of P(S) are sets. Out of the 32 subsets of S (which form P(S)), we are just removing one (S). So there should be 31 subset left in P(S)-S (instead of 32).

Answer 35

no @mathmate you're mixing sets with elements too.

Answer 36

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

Answer 37

Elements of a set can be a set also. A power set has 2^n elements, each of which is a subset of S. That is why there are 32 subsets of S contained in P(S), as jim_thompson5910 has enumerated.
What is inside of each subset is not relevant for the moment. Removing S which is a subset of P(S) does not remove the elements of S from the other subsets. Each subset is encapsulated and is not "visible". The only elements we see are subsets.

Answer 38

Yeah but why are you removing S?

Answer 39

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

Answer 40

You must remove all the elements that are in \(\wp S \) and in \(S\) but I can't find those elements.

Answer 41

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

Answer 42

no, you're removing everything actually

the elements 1, 2, 3, 4, 5 are all in S and each number is found somewhere in P(S)

so that's why I'm thinking your original answer of the empty set is correct

Answer 43

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

Answer 44

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

Answer 45

true, not thinking on this one lol

Answer 46

lol no problem.

Answer 47

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

Answer 48

yes i do

Answer 49

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

Answer 50

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

Answer 51

For the original question you provided, the relationship holds. Because your original set contains non-set elements, then the power set contains only sets. Thus the two have nothing in common.

Generally speaking though this relationship doesn't seem to hold.

Answer 52

probably this, but I'm not 100% sure

P(S) =
{
{1,{1}}
{1}, {{1}}
{}
}

Answer 53

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

Answer 54

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

Answer 55

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

Answer 56

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

Answer 57

Ummm, I'm pretty sure the power set is the set of all subsets. If you take out the set itself, it would be the set of all proper subsets.
So if \(S = \{1, 2\}\) then \(\wp S - S = \{\emptyset , \{1\}, \{2\}\}\)

Answer 58

I was thinking along the same lines

Answer 59

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

Answer 60

This problem is tricky

Answer 61

Ohhh, I read it wrong then.

Answer 62

So basically, it's just the power set then, because the power set does NOT have any elements of S in it.

Answer 63

Except perhaps the empty set?

Answer 64

yeah exactly !

Answer 65

not, the empty set is not in S

Answer 66

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

Answer 67

mmm ok.

Answer 68

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

Answer 69

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

Answer 70

The power set does not contain any of the elements 1,2,3,4, nor 5. It contains the subsets {1}, {2}, ...{5}, {1,2}, ...{1,2,3,4,5}.
It's the last item that we are removing, and only that.

Answer 71

But you have no reason to remove the last item @mathmate

Answer 72

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

Answer 73

Because \(S\not \in S\)

Answer 74

S is not a proper subset of S, but it is a subset of S.
Also, S is a subset of P(S), because P(S) contains 32 sets, one of which is S (the only one that has 5 elements).

Answer 75

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

Answer 76

The elements of the power set are:
P(S)={ empty-set, {1},{2},{3},{4},{5},{1,2},{1,3},{1,4},{1,5},
{2,3},{2,4},{2,5},{3,4},{3,5},{4,5},{1,2,3},{1,2,4},{1,2,5},
{2,3,4},{2,3,5},{3,4,5},{1,2,3,4},{1,2,3,5},{2,3,4,5},
{1,2,3,4,5} }
The last element is S.

Answer 77

Ok. now list all the elements of S.

Answer 78

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

Answer 79

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

Answer 80

brb

Answer 81

ok

Answer 82

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

Answer 83

* 31 elements.

Answer 84

Why you're removing the set S?

Answer 85

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

Answer 86

I thought that was the original question:
"Given the following set \(S:=\{1,2,3,4,5\}\) evaluate the following expression:
\[ P(S) - S\] "
(I edited P(S) )

Answer 87

Yes that is the question. But I think you're not applying correctly the definition of set difference. You're mixing up elements with sets.

Answer 88

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

Answer 89

Sets contain elements. A school can have 6 classrooms, each of which is a set of 30 students.

So the school has 6 elements, each is a set of 30 students.
If the elements of the subsets are "let out" of the subsets, then we will have a lot of duplicates. A set cannot have duplicate elements. So we must treat the elements of P(S) as sets, which in turn is made up of their own elements 1,2,3...
All the 32 elements of P(S) have been enumerated, and each itself is a set (or subset of S).

Answer 90

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

Answer 91

So elements cannot be subsets since they are not sets to begin with. 1 is not a subset of {1, 2}, for example?

Answer 92

Is 1 a subset of {1, 2} ?

Answer 93

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

Answer 94

I always thought subsets were sets whose elements were contained in another set. By this definition, elements cannot be subsets unless they are sets.

Answer 95

All the elements of the power set are subsets, which are different from the element(s) they contain. Example, 1 is not equal to {1}.
For example, the power set of {1} is {empty-set, {1}}
1 does not have a power set.

Answer 96

yeah you're right @LogicalApple

Answer 97

So you agree with this @mathmate \[\wp S - S = \wp S\]