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

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NodataBest ResponseYou've already chosen the best response.1
I think that the result is the empty set.
 one year ago

NodataBest ResponseYou've already chosen the best response.1
\[\wp S  S = \emptyset\] because there is not an element on S that equals an elemento of the power set of S.
 one year ago

NodataBest ResponseYou've already chosen the best response.1
\(\not \exists x\in S\;\; x\in\wp S \)
 one year ago

NodataBest ResponseYou've already chosen the best response.1
@brinethery what do you think?
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.0
If \[\wp S\] is the power set, then list out everything in the power set but kick out the set S
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.0
ok, but you won't get the empty set
 one year ago

NodataBest ResponseYou've already chosen the best response.1
But I didn't kick out the set S.
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.0
that's the what the "  S " portion is saying
 one year ago

NodataBest ResponseYou've already chosen the best response.1
What I think it says is that I must kick out the elements of S that are elements of P(S)
 one year ago

NodataBest ResponseYou've already chosen the best response.1
and none of the elements of S are in P(S)
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.0
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} {} }
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.0
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} {} }
 one year ago

NodataBest ResponseYou've already chosen the best response.1
\(1\not \in S\) for example
 one year ago

NodataBest ResponseYou've already chosen the best response.1
because \(1\not = \{1\}\)
 one year ago

NodataBest ResponseYou've already chosen the best response.1
And the same happens with 2,3,4,5
 one year ago

NodataBest ResponseYou've already chosen the best response.1
Mmm then I think \(\wp S  S = \wp S\)
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.0
not quite, P(S)  S does not equal P(S)
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.0
close though, because you're only erasing one element of the powerset
 one year ago

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

jim_thompson5910Best 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)
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.0
P(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} }
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.0
heres a page on the power set http://en.wikipedia.org/wiki/Power_set
 one year ago

NodataBest ResponseYou've already chosen the best response.1
I'm agree with the power set expansion. But I think you're difference is wrong
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.0
P(S)  S means "start with P(S), then remove the set S itself since S is an element of P(S)"
 one year ago

NodataBest ResponseYou've already chosen the best response.1
\(BA=\{x\in B  x\not \in A\}\)
 one year ago

NodataBest ResponseYou've already chosen the best response.1
What are you saying is this: \(\wp S  \{S\} \)
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.0
oh good point, I'm mixing up sets and elements here...
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.0
so P(S)  S would be start with P(S) and kick out any elements you find in S
 one year ago

NodataBest ResponseYou've already chosen the best response.1
Yeah, and there is no elements that are in both sets.
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.0
so then your original answer of empty set works
 one year ago

NodataBest ResponseYou've already chosen the best response.1
No, I think the right answer is \(\wp S  S = \wp S\) because there is no elements to quit from \(\wp S\)
 one year ago

mathmateBest ResponseYou've already chosen the best response.1
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).
 one year ago

NodataBest ResponseYou've already chosen the best response.1
no @mathmate you're mixing sets with elements too.
 one year ago

NodataBest ResponseYou've already chosen the best response.1
you can't quit \(S\) because \(S\not \in S\)
 one year ago

mathmateBest ResponseYou've already chosen the best response.1
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.
 one year ago

NodataBest ResponseYou've already chosen the best response.1
Yeah but why are you removing S?
 one year ago

NodataBest ResponseYou've already chosen the best response.1
When you do a set difference you remove the elements that both sets have in common right?
 one year ago

NodataBest ResponseYou've already chosen the best response.1
You must remove all the elements that are in \(\wp S \) and in \(S\) but I can't find those elements.
 one year ago

NodataBest ResponseYou've already chosen the best response.1
So you're removing nothing from \(\wp S\)
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.0
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
 one year ago

NodataBest ResponseYou've already chosen the best response.1
You can't remove 1 because this element is not in \(\wp S\)
 one year ago

NodataBest ResponseYou've already chosen the best response.1
\(1\not = \{1\}\) for example
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.0
true, not thinking on this one lol
 one year ago

NodataBest ResponseYou've already chosen the best response.1
So do you agree with my answer @jim_thompson5910 ?\[\wp S  S = \wp S\]
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.0
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
 one year ago

LogicalAppleBest ResponseYou've already chosen the best response.1
What if A = {1, {1}}, what would its power set look like?
 one year ago

LogicalAppleBest ResponseYou've already chosen the best response.1
For 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.
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.0
probably this, but I'm not 100% sure P(S) = { {1,{1}} {1}, {{1}} {} }
 one year ago

NodataBest ResponseYou've already chosen the best response.1
\[\wp A = \{\emptyset, \{1\}, \{\{1\}\}, A\}\]
 one year ago

LogicalAppleBest ResponseYou've already chosen the best response.1
In that instance, the original set contains the element {1}, but so does the power set.
 one year ago

NodataBest ResponseYou've already chosen the best response.1
Yes you're right @LogicalApple that relation holds for sets with nonset elements.
 one year ago

NodataBest ResponseYou've already chosen the best response.1
I mean \(\wp A  A \not = \wp A\)
 one year ago

wioBest ResponseYou've already chosen the best response.1
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 \(\wp S  S = \{\emptyset , \{1\}, \{2\}\}\)
 one year ago

swissgirlBest ResponseYou've already chosen the best response.0
I was thinking along the same lines
 one year ago

NodataBest ResponseYou've already chosen the best response.1
You can take out the set itself if you have the following expression: \[\wp S  \{S\}\]
 one year ago

NodataBest ResponseYou've already chosen the best response.1
This problem is tricky
 one year ago

wioBest ResponseYou've already chosen the best response.1
Ohhh, I read it wrong then.
 one year ago

wioBest ResponseYou've already chosen the best response.1
So basically, it's just the power set then, because the power set does NOT have any elements of S in it.
 one year ago

wioBest ResponseYou've already chosen the best response.1
Except perhaps the empty set?
 one year ago

NodataBest ResponseYou've already chosen the best response.1
not, the empty set is not in S
 one year ago

wioBest ResponseYou've already chosen the best response.1
Yeah, in this case it there isn't any elements in common, but I was just thinking generally.
 one year ago

wioBest ResponseYou've already chosen the best response.1
I'm used to seeing the power set denoted with \mathcal P : \(\mathcal P(S) \)
 one year ago

NodataBest ResponseYou've already chosen the best response.1
Yeah there are several way of writting it. I just use the same as the one in my book hehe.
 one year ago

mathmateBest ResponseYou've already chosen the best response.1
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.
 one year ago

NodataBest ResponseYou've already chosen the best response.1
But you have no reason to remove the last item @mathmate
 one year ago

mathmateBest ResponseYou've already chosen the best response.1
Not at all, but that's what the question wants.
 one year ago

NodataBest ResponseYou've already chosen the best response.1
Because \(S\not \in S\)
 one year ago

mathmateBest ResponseYou've already chosen the best response.1
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).
 one year ago

NodataBest ResponseYou've already chosen the best response.1
I agree with \(S\subseteq\wp S\).
 one year ago

mathmateBest ResponseYou've already chosen the best response.1
The 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.
 one year ago

NodataBest ResponseYou've already chosen the best response.1
Ok. now list all the elements of S.
 one year ago

NodataBest ResponseYou've already chosen the best response.1
And quit the elements that are in S and the power set of S
 one year ago

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

mathmateBest ResponseYou've already chosen the best response.1
So P(S)\S will be left with only 31 subsets, the first 31.
 one year ago

NodataBest ResponseYou've already chosen the best response.1
Why you're removing the set S?
 one year ago

NodataBest ResponseYou've already chosen the best response.1
S is not an element of S, so you can't remove it.
 one year ago

mathmateBest ResponseYou've already chosen the best response.1
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) )
 one year ago

NodataBest ResponseYou've already chosen the best response.1
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.
 one year ago

NodataBest ResponseYou've already chosen the best response.1
You must remove from the power set of S the elements of S that are in the power set of S too.
 one year ago

mathmateBest ResponseYou've already chosen the best response.1
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).
 one year ago

NodataBest ResponseYou've already chosen the best response.1
yeas. the power set of S has set elements. then you can't remove non set element from it.
 one year ago

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

LogicalAppleBest ResponseYou've already chosen the best response.1
Is 1 a subset of {1, 2} ?
 one year ago

NodataBest ResponseYou've already chosen the best response.1
1 is not a set. so it can't be a subset.
 one year ago

LogicalAppleBest ResponseYou've already chosen the best response.1
I always thought subsets were sets whose elements were contained in another set. By this definition, elements cannot be subsets unless they are sets.
 one year ago

mathmateBest ResponseYou've already chosen the best response.1
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 {emptyset, {1}} 1 does not have a power set.
 one year ago

NodataBest ResponseYou've already chosen the best response.1
yeah you're right @LogicalApple
 one year ago

NodataBest ResponseYou've already chosen the best response.1
So you agree with this @mathmate \[\wp S  S = \wp S\]
 one year ago

mathmateBest ResponseYou've already chosen the best response.1
By 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.
 one year ago

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

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

mathmateBest ResponseYou've already chosen the best response.1
In fact, in general, P(S)S has 2^n1 elements.
 one year ago

NodataBest ResponseYou've already chosen the best response.1
Remember the definition of a set difference: \[AB=\{x\in Ax\not \in B\}\]
 one year ago

NodataBest ResponseYou've already chosen the best response.1
I'm not looking for the number of elements. but for a resulting set.
 one year ago

mathmateBest ResponseYou've already chosen the best response.1
Actually 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} \( \notin S \)
 one year ago

LogicalAppleBest ResponseYou've already chosen the best response.1
Should 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.
 one year ago

NodataBest ResponseYou've already chosen the best response.1
we can't exclude the empty set and the original set. @LogicalApple but I can't find I way to explain it to @mathmate
 one year ago

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

LogicalAppleBest ResponseYou've already chosen the best response.1
The power set was generated by wolframalpha
 one year ago

mathmateBest ResponseYou've already chosen the best response.1
Note that 1 is an element of S, but {1} is not, it is a subset of S.
 one year ago

LogicalAppleBest ResponseYou've already chosen the best response.1
Right. Because all elements of S are nonsets, then S cannot share any elements in common with P since P is made up of sets.
 one year ago

LogicalAppleBest ResponseYou've already chosen the best response.1
*P also has {}, but S does not.
 one year ago

NodataBest ResponseYou've already chosen the best response.1
Great job @LogicalApple Thank you!
 one year ago

NodataBest ResponseYou've already chosen the best response.1
by the way the empty set is a subset of S but is not an element of S @mathmate
 one year ago

wioBest ResponseYou've already chosen the best response.1
I don't understand what the problem here is... \[ \mathcal P(S)  S = \mathcal P(s) \]
 one year ago

LogicalAppleBest ResponseYou've already chosen the best response.1
@wio I have a tough problem ready for you later though..
 one year ago

NodataBest ResponseYou've already chosen the best response.1
that expression hold for \(S = \{1,2,3,4,5\}\)
 one year ago

LogicalAppleBest ResponseYou've already chosen the best response.1
I would think that expression holds unless S contains an element that is also a subset of S.
 one year ago

wioBest ResponseYou've already chosen the best response.1
It holds when: \[ \neg \exists A \subseteq S \wedge A \in S \] Why is this question not closed?
 one year ago

NodataBest ResponseYou've already chosen the best response.1
It is actually closed haha.
 one year ago

wioBest ResponseYou've already chosen the best response.1
I mean, why so many responses still?
 one year ago

NodataBest ResponseYou've already chosen the best response.1
Yeah because @mathmate still negates our answer hehe.
 one year ago

mathmateBest 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.
 one year ago

mathmateBest ResponseYou've already chosen the best response.1
Therefore, by definition of \ , {} gets kicked out of P(S)S.
 one year ago

mathmateBest ResponseYou've already chosen the best response.1
Sorry 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.
 one year ago

mathmateBest ResponseYou've already chosen the best response.1
@Nodata What is your answer by the way, I already presented mine, with 30 elements.
 one year ago

NodataBest ResponseYou've already chosen the best response.1
The empty set can't be removed because is not an element of S neither S.
 one year ago

NodataBest ResponseYou've already chosen the best response.1
Look a this its more clear.
 one year ago

mathmateBest ResponseYou've already chosen the best response.1
One 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} }
 one year ago

NodataBest ResponseYou've already chosen the best response.1
The fact that the empty set is a subset does not imply that the empty set is an element of every set.
 one year ago

NodataBest ResponseYou've already chosen the best response.1
For example \( \emptyset \subset S\) but \(\emptyset \not \in S \)
 one year ago

mathmateBest ResponseYou've already chosen the best response.1
Absolutely 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.
 one year ago

NodataBest ResponseYou've already chosen the best response.1
Then it's not correct to remove the empty set from the powerset in this case right?
 one year ago

mathmateBest ResponseYou've already chosen the best response.1
However, 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.
 one year ago

NodataBest ResponseYou've already chosen the best response.1
No the empty set is not in every set. The empty set is a subset of every set.
 one year ago

mathmateBest ResponseYou've already chosen the best response.1
Yes, 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.
 one year ago
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