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

This Question is Closed

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I think that the result is the empty set.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\wp S  S = \emptyset\] because there is not an element on S that equals an elemento of the power set of S.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\(\not \exists x\in S\;\; x\in\wp S \)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@brinethery what do you think?

jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.0If \[\wp S\] is the power set, then list out everything in the power set but kick out the set S

jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.0ok, but you won't get the empty set

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0But I didn't kick out the set S.

jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.0that's the what the "  S " portion is saying

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0What I think it says is that I must kick out the elements of S that are elements of P(S)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and none of the elements of S are in P(S)

jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.0If 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} {} }

jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.0you 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} {} }

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\(1\not \in S\) for example

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0because \(1\not = \{1\}\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0And the same happens with 2,3,4,5

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Mmm then I think \(\wp S  S = \wp S\)

jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.0not quite, P(S)  S does not equal P(S)

jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.0close though, because you're only erasing one element of the powerset

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I cant quit {1,2,3,4,5} because \(\{1,2,3,4,5\}\not \in S\)

jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.0{1,2,3,4,5} is S, so it's a subset of P(S) S is a subset of P(S)

jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.0P(S) = { all possible subsets of S } P(S) = { S, {sets of size n1 where each element is in S} {sets of size n2 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} }

jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.0heres a page on the power set http://en.wikipedia.org/wiki/Power_set

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I'm agree with the power set expansion. But I think you're difference is wrong

jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.0P(S)  S means "start with P(S), then remove the set S itself since S is an element of P(S)"

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\(BA=\{x\in B  x\not \in A\}\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0What are you saying is this: \(\wp S  \{S\} \)

jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.0oh good point, I'm mixing up sets and elements here...

jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.0so P(S)  S would be start with P(S) and kick out any elements you find in S

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yeah, and there is no elements that are in both sets.

jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.0so then your original answer of empty set works

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0No, I think the right answer is \(\wp S  S = \wp S\) because there is no elements to quit from \(\wp S\)

mathmate
 3 years ago
Best ResponseYou've already chosen the best response.1We 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).

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0no @mathmate you're mixing sets with elements too.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you can't quit \(S\) because \(S\not \in S\)

mathmate
 3 years ago
Best ResponseYou've already chosen the best response.1Elements 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yeah but why are you removing S?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0When you do a set difference you remove the elements that both sets have in common right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You must remove all the elements that are in \(\wp S \) and in \(S\) but I can't find those elements.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So you're removing nothing from \(\wp S\)

jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.0no, 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You can't remove 1 because this element is not in \(\wp S\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\(1\not = \{1\}\) for example

jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.0true, not thinking on this one lol

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So do you agree with my answer @jim_thompson5910 ?\[\wp S  S = \wp S\]

jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.0there'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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0What if A = {1, {1}}, what would its power set look like?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0For the original question you provided, the relationship holds. Because your original set contains nonset 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.

jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.0probably this, but I'm not 100% sure P(S) = { {1,{1}} {1}, {{1}} {} }

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\wp A = \{\emptyset, \{1\}, \{\{1\}\}, A\}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0In that instance, the original set contains the element {1}, but so does the power set.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yes you're right @LogicalApple that relation holds for sets with nonset elements.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I mean \(\wp A  A \not = \wp A\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Ummm, 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 \(\wp S  S = \{\emptyset , \{1\}, \{2\}\}\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I was thinking along the same lines

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You can take out the set itself if you have the following expression: \[\wp S  \{S\}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0This problem is tricky

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Ohhh, I read it wrong then.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So basically, it's just the power set then, because the power set does NOT have any elements of S in it.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Except perhaps the empty set?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0not, the empty set is not in S

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yeah, in this case it there isn't any elements in common, but I was just thinking generally.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I'm used to seeing the power set denoted with \mathcal P : \(\mathcal P(S) \)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yeah there are several way of writting it. I just use the same as the one in my book hehe.

mathmate
 3 years ago
Best ResponseYou've already chosen the best response.1The 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0But you have no reason to remove the last item @mathmate

mathmate
 3 years ago
Best ResponseYou've already chosen the best response.1Not at all, but that's what the question wants.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Because \(S\not \in S\)

mathmate
 3 years ago
Best ResponseYou've already chosen the best response.1S 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).

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I agree with \(S\subseteq\wp S\).

mathmate
 3 years ago
Best ResponseYou've already chosen the best response.1The elements of the power set are: P(S)={ emptyset, {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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Ok. now list all the elements of S.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0And quit the elements that are in S and the power set of S

mathmate
 3 years ago
Best ResponseYou've already chosen the best response.1Note that P(S) does not have any element like 1,2,3,... , only subsets.

mathmate
 3 years ago
Best ResponseYou've already chosen the best response.1So P(S)\S will be left with only 31 subsets, the first 31.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Why you're removing the set S?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0S is not an element of S, so you can't remove it.

mathmate
 3 years ago
Best ResponseYou've already chosen the best response.1I 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) )

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yes that is the question. But I think you're not applying correctly the definition of set difference. You're mixing up elements with sets.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You must remove from the power set of S the elements of S that are in the power set of S too.

mathmate
 3 years ago
Best ResponseYou've already chosen the best response.1Sets 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).

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yeas. the power set of S has set elements. then you can't remove non set element from it.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So elements cannot be subsets since they are not sets to begin with. 1 is not a subset of {1, 2}, for example?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Is 1 a subset of {1, 2} ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.01 is not a set. so it can't be a subset.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I always thought subsets were sets whose elements were contained in another set. By this definition, elements cannot be subsets unless they are sets.

mathmate
 3 years ago
Best ResponseYou've already chosen the best response.1All 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 {emptyset, {1}} 1 does not have a power set.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yeah you're right @LogicalApple

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So you agree with this @mathmate \[\wp S  S = \wp S\]

mathmate
 3 years ago
Best ResponseYou've already chosen the best response.1By definition, the power set cannot contain elements. That does not stop us from making a set which contains: X={1,2,3,{1},{2},{3}} which also demonstrates the difference between a subset and an element.

mathmate
 3 years ago
Best ResponseYou've already chosen the best response.1Sorry, I have a different opinion. P(S) has 32 elements (which are subsets), while P(S)S has only 31.

mathmate
 3 years ago
Best ResponseYou've already chosen the best response.1I meant to say in the previous post: "By definition, the power set cannot contain elements of the original set".

mathmate
 3 years ago
Best ResponseYou've already chosen the best response.1In fact, in general, P(S)S has 2^n1 elements.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Remember the definition of a set difference: \[AB=\{x\in Ax\not \in B\}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I'm not looking for the number of elements. but for a resulting set.

mathmate
 3 years ago
Best ResponseYou've already chosen the best response.1Actually you have a point. P(S)S does not have 31 elements, but only 30, because we have to exclude the empty set AND S, which leaves P(S)S = { {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} } with a cardinality of 30. Note that \( 1 \in S \) but {1} \( \notin S \)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Should we really exclude the empty set, {}, and the original set, {1, 2, 3, 4, 5}? These are both elements, x, of P such that x is not also in S. In fact, all elements of P are elements that are not also in S. At least in this instance.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0we can't exclude the empty set and the original set. @LogicalApple but I can't find I way to explain it to @mathmate

mathmate
 3 years ago
Best ResponseYou've already chosen the best response.1empty set belongs to both P(S) and S, but not 1,2,3...5, so these don't need to be excluded.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The power set was generated by wolframalpha

mathmate
 3 years ago
Best ResponseYou've already chosen the best response.1Note that 1 is an element of S, but {1} is not, it is a subset of S.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Right. Because all elements of S are nonsets, then S cannot share any elements in common with P since P is made up of sets.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0*P also has {}, but S does not.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Great job @LogicalApple Thank you!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0by the way the empty set is a subset of S but is not an element of S @mathmate

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I don't understand what the problem here is... \[ \mathcal P(S)  S = \mathcal P(s) \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@wio I have a tough problem ready for you later though..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0that expression hold for \(S = \{1,2,3,4,5\}\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I would think that expression holds unless S contains an element that is also a subset of S.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0It holds when: \[ \neg \exists A \subseteq S \wedge A \in S \] Why is this question not closed?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0It is actually closed haha.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I mean, why so many responses still?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yeah because @mathmate still negates our answer hehe.

mathmate
 3 years ago
Best ResponseYou've already chosen the best response.1{} is a subset of S, that's how it got into P(S)! In fact, {} is a subset of every set.

mathmate
 3 years ago
Best ResponseYou've already chosen the best response.1Therefore, by definition of \ , {} gets kicked out of P(S)S.

mathmate
 3 years ago
Best ResponseYou've already chosen the best response.1Sorry that we don't come to a consensus. We have all tried our best, and I think it is best to leave it at that.

mathmate
 3 years ago
Best ResponseYou've already chosen the best response.1@Nodata What is your answer by the way, I already presented mine, with 30 elements.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The empty set can't be removed because is not an element of S neither S.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Look a this its more clear.

mathmate
 3 years ago
Best ResponseYou've already chosen the best response.1One of the properties of the empty set is: "For any set A: The empty set is a subset of A:" Using your definition of A−B={x∈Ax∉B} since {} belongs to both P(S) and S, by definition, {} should not be in AB. However, I just realize that {} is present in every set, even if we exclude it. So you're right, P(S) contains 31 subsets, including {}, P(S)S={ emptyset, {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},{2,3,4,5} }

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The fact that the empty set is a subset does not imply that the empty set is an element of every set.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0For example \( \emptyset \subset S\) but \(\emptyset \not \in S \)

mathmate
 3 years ago
Best ResponseYou've already chosen the best response.1Absolutely true. The empty set IS a \( subset \) of every set. It is an element of the power set, but not an \(element\) of every set.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Then it's not correct to remove the empty set from the powerset in this case right?

mathmate
 3 years ago
Best ResponseYou've already chosen the best response.1However, I just realize that {} is present in every set, even if we exclude it. should have been worded However, I just realize that {} is a subset of every set, even if we exclude it. Also, since it is NOT an element of S, we do not have to exclude it.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0No the empty set is not in every set. The empty set is a subset of every set.

mathmate
 3 years ago
Best ResponseYou've already chosen the best response.1Yes, I agree that P(S)S has 31 subsets (elements), including {}. {} is a subset of every set, but is not an element of every set, notably it is not an element of S, so it does not need to be removed from P(S)S.
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