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

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

Mathematics
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I think that the result is the empty set.
\[\wp S - S = \emptyset\] because there is not an element on S that equals an elemento of the power set of S.
\(\not \exists x\in S\;|\; x\in\wp S \)

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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)
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} {} }
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} {} }
\(1\not \in S\) 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)
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} }
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
No, I think the right answer is \(\wp S - S = \wp S\) because there is no elements to quit from \(\wp S\)
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).
no @mathmate you're mixing sets with elements too.
you can't quit \(S\) because \(S\not \in S\)
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.
Yeah but why are you removing S?
When you do a set difference you remove the elements that both sets have in common right?
You must remove all the elements that are in \(\wp S \) and in \(S\) but I can't find those elements.
So you're removing nothing from \(\wp S\)
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
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?
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.
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 \(\wp A - A \not = \wp A\)
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 \(\wp S - S = \{\emptyset , \{1\}, \{2\}\}\)
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.
So basically, it's just the power set then, because the power set does NOT have any elements of S in it.
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.
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.
But you have no reason to remove the last item @mathmate
Not at all, but that's what the question wants.
Because \(S\not \in S\)
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).
I agree with \(S\subseteq\wp S\).
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.
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.
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) )
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.
You must remove from the power set of S the elements of S that are in the power set of S too.
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).
yeas. the power set of S has set elements. then you can't remove non set element from it.
So elements cannot be subsets since they are not sets to begin with. 1 is not a subset of {1, 2}, for example?
Is 1 a subset of {1, 2} ?
1 is not a set. so it can't be a subset.
I always thought subsets were sets whose elements were contained in another set. By this definition, elements cannot be subsets unless they are sets.
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.
yeah you're right @LogicalApple
So you agree with this @mathmate \[\wp S - S = \wp S\]
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.
Sorry, I have a different opinion. P(S) has 32 elements (which are subsets), while P(S)-S has only 31.
I meant to say in the previous post: "By definition, the power set cannot contain elements of the original set".
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.
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} \( \notin S \)
brb
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.
we can't exclude the empty set and the original set. @LogicalApple but I can't find I way to explain it to @mathmate
empty set belongs to both P(S) and S, but not 1,2,3...5, so these don't need to be excluded.
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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.
Right. Because all elements of S are non-sets, then S cannot share any elements in common with P since P is made up of sets.
*P also has {}, but S does not.
Great job @LogicalApple Thank you!
by the way the empty set is a subset of S but is not an element of S @mathmate
I don't understand what the problem here is... \[ \mathcal P(S) - S = \mathcal P(s) \]
@wio I have a tough problem ready for you later though..
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?
Yeah because @mathmate still negates our answer hehe.
{} 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.
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.
@No-data What is your answer by the way, I already presented mine, with 30 elements.
The empty set can't be removed because is not an element of S neither S.
Look a this its more clear.
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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∈A|x∉B} since {} belongs to both P(S) and S, by definition, {} should not be in A-B. 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={ 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},{2,3,4,5} }
The fact that the empty set is a subset does not imply that the empty set is an element of every set.
For example \( \emptyset \subset S\) but \(\emptyset \not \in S \)
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.
Then it's not correct to remove the empty set from the powerset in this case right?
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.
No the empty set is not in every set. The empty set is a subset of every set.
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.

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