## No-data Group Title Given the following set $$S:=\{1,2,3,4,5\}$$ evaluate the following expression: $\wp S - S$ one year ago one year ago

1. No-data Group Title

I think that the result is the empty set.

2. No-data Group Title

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

3. No-data Group Title

$$\not \exists x\in S\;|\; x\in\wp S$$

4. No-data Group Title

Am I right?

5. No-data Group Title

@brinethery what do you think?

6. jim_thompson5910 Group Title

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

7. No-data Group Title

That is what I did.

8. jim_thompson5910 Group Title

ok, but you won't get the empty set

9. No-data Group Title

But I didn't kick out the set S.

10. jim_thompson5910 Group Title

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

11. No-data Group Title

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

12. No-data Group Title

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

13. jim_thompson5910 Group Title

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} {} }

14. jim_thompson5910 Group Title

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} {} }

15. No-data Group Title

$$1\not \in S$$ for example

16. No-data Group Title

because $$1\not = \{1\}$$

17. No-data Group Title

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

18. No-data Group Title

Mmm then I think $$\wp S - S = \wp S$$

19. jim_thompson5910 Group Title

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

20. jim_thompson5910 Group Title

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

21. No-data Group Title

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

22. jim_thompson5910 Group Title

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

23. jim_thompson5910 Group Title

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} }

24. jim_thompson5910 Group Title

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

25. No-data Group Title

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

26. jim_thompson5910 Group Title

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

27. No-data Group Title

$$B-A=\{x\in B | x\not \in A\}$$

28. No-data Group Title

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

29. jim_thompson5910 Group Title

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

30. jim_thompson5910 Group Title

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

31. No-data Group Title

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

32. jim_thompson5910 Group Title

33. No-data Group Title

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

34. mathmate Group Title

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).

35. No-data Group Title

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

36. No-data Group Title

you can't quit $$S$$ because $$S\not \in S$$

37. mathmate Group Title

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.

38. No-data Group Title

Yeah but why are you removing S?

39. No-data Group Title

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

40. No-data Group Title

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

41. No-data Group Title

So you're removing nothing from $$\wp S$$

42. jim_thompson5910 Group Title

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

43. No-data Group Title

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

44. No-data Group Title

$$1\not = \{1\}$$ for example

45. jim_thompson5910 Group Title

true, not thinking on this one lol

46. No-data Group Title

lol no problem.

47. No-data Group Title

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

48. jim_thompson5910 Group Title

yes i do

49. jim_thompson5910 Group Title

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

50. LogicalApple Group Title

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

51. LogicalApple Group Title

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.

52. jim_thompson5910 Group Title

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

53. No-data Group Title

$\wp A = \{\emptyset, \{1\}, \{\{1\}\}, A\}$

54. LogicalApple Group Title

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

55. No-data Group Title

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

56. No-data Group Title

I mean $$\wp A - A \not = \wp A$$

57. wio Group Title

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 $$\wp S - S = \{\emptyset , \{1\}, \{2\}\}$$

58. swissgirl Group Title

I was thinking along the same lines

59. No-data Group Title

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

60. No-data Group Title

This problem is tricky

61. wio Group Title

Ohhh, I read it wrong then.

62. wio Group Title

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

63. wio Group Title

Except perhaps the empty set?

64. No-data Group Title

yeah exactly !

65. No-data Group Title

not, the empty set is not in S

66. wio Group Title

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

67. No-data Group Title

mmm ok.

68. wio Group Title

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

69. No-data Group Title

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

70. mathmate Group Title

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.

71. No-data Group Title

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

72. mathmate Group Title

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

73. No-data Group Title

Because $$S\not \in S$$

74. mathmate Group Title

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).

75. No-data Group Title

I agree with $$S\subseteq\wp S$$.

76. mathmate Group Title

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.

77. No-data Group Title

Ok. now list all the elements of S.

78. No-data Group Title

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

79. mathmate Group Title

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

80. mathmate Group Title

brb

81. No-data Group Title

ok

82. mathmate Group Title

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

83. mathmate Group Title

* 31 elements.

84. No-data Group Title

Why you're removing the set S?

85. No-data Group Title

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

86. mathmate Group Title

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) )

87. No-data Group Title

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.

88. No-data Group Title

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

89. mathmate Group Title

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).

90. No-data Group Title

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

91. LogicalApple Group Title

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

92. LogicalApple Group Title

Is 1 a subset of {1, 2} ?

93. No-data Group Title

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

94. LogicalApple Group Title

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

95. mathmate Group Title

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.

96. No-data Group Title

yeah you're right @LogicalApple

97. No-data Group Title

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

98. mathmate Group Title

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.

99. mathmate Group Title

Sorry, I have a different opinion. P(S) has 32 elements (which are subsets), while P(S)-S has only 31.

100. mathmate Group Title

I meant to say in the previous post: "By definition, the power set cannot contain elements of the original set".

101. mathmate Group Title

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

102. No-data Group Title

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

103. No-data Group Title

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

104. mathmate Group Title

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} $$\notin S$$

105. mathmate Group Title

brb

106. LogicalApple Group Title

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.

107. No-data Group Title

we can't exclude the empty set and the original set. @LogicalApple but I can't find I way to explain it to @mathmate

108. mathmate Group Title

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

109. LogicalApple Group Title

110. LogicalApple Group Title

The power set was generated by wolframalpha

111. mathmate Group Title

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

112. LogicalApple Group Title

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.

113. LogicalApple Group Title

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

114. No-data Group Title

Great job @LogicalApple Thank you!

115. No-data Group Title

by the way the empty set is a subset of S but is not an element of S @mathmate

116. wio Group Title

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

117. LogicalApple Group Title

@wio I have a tough problem ready for you later though..

118. No-data Group Title

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

119. wio Group Title

Yeah.

120. LogicalApple Group Title

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

121. wio Group Title

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

122. No-data Group Title

It is actually closed haha.

123. wio Group Title

I mean, why so many responses still?

124. No-data Group Title

Yeah because @mathmate still negates our answer hehe.

125. mathmate Group Title

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

126. mathmate Group Title

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

127. mathmate Group Title

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.

128. mathmate Group Title

129. No-data Group Title

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

130. No-data Group Title

Look a this its more clear.

131. mathmate Group Title

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} }

132. No-data Group Title

The fact that the empty set is a subset does not imply that the empty set is an element of every set.

133. No-data Group Title

For example $$\emptyset \subset S$$ but $$\emptyset \not \in S$$

134. mathmate Group Title

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.

135. No-data Group Title

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

136. mathmate Group Title

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.

137. No-data Group Title

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

138. mathmate Group Title

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.