## anonymous 4 years ago Given the following set $$S:=\{1,2,3,4,5\}$$ evaluate the following expression: $\wp S - S$

1. anonymous

I think that the result is the empty set.

2. anonymous

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

3. anonymous

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

4. anonymous

Am I right?

5. anonymous

@brinethery what do you think?

6. jim_thompson5910

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

7. anonymous

That is what I did.

8. jim_thompson5910

ok, but you won't get the empty set

9. anonymous

But I didn't kick out the set S.

10. jim_thompson5910

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

11. anonymous

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

12. anonymous

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

13. jim_thompson5910

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

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. anonymous

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

16. anonymous

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

17. anonymous

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

18. anonymous

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

19. jim_thompson5910

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

20. jim_thompson5910

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

21. anonymous

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

22. jim_thompson5910

{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

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

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

25. anonymous

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

26. jim_thompson5910

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

27. anonymous

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

28. anonymous

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

29. jim_thompson5910

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

30. jim_thompson5910

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

31. anonymous

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

32. jim_thompson5910

so then your original answer of empty set works

33. anonymous

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

34. mathmate

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. anonymous

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

36. anonymous

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

37. mathmate

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. anonymous

Yeah but why are you removing S?

39. anonymous

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

40. anonymous

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

41. anonymous

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

42. jim_thompson5910

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. anonymous

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

44. anonymous

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

45. jim_thompson5910

true, not thinking on this one lol

46. anonymous

lol no problem.

47. anonymous

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

48. jim_thompson5910

yes i do

49. jim_thompson5910

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. anonymous

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

51. anonymous

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

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

53. anonymous

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

54. anonymous

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

55. anonymous

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

56. anonymous

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

57. anonymous

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

I was thinking along the same lines

59. anonymous

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

60. anonymous

This problem is tricky

61. anonymous

Ohhh, I read it wrong then.

62. anonymous

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

63. anonymous

Except perhaps the empty set?

64. anonymous

yeah exactly !

65. anonymous

not, the empty set is not in S

66. anonymous

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

67. anonymous

mmm ok.

68. anonymous

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

69. anonymous

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

70. mathmate

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. anonymous

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

72. mathmate

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

73. anonymous

Because $$S\not \in S$$

74. mathmate

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. anonymous

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

76. mathmate

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. anonymous

Ok. now list all the elements of S.

78. anonymous

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

79. mathmate

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

80. mathmate

brb

81. anonymous

ok

82. mathmate

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

83. mathmate

* 31 elements.

84. anonymous

Why you're removing the set S?

85. anonymous

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

86. mathmate

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. anonymous

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. anonymous

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

89. mathmate

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. anonymous

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

91. anonymous

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

92. anonymous

Is 1 a subset of {1, 2} ?

93. anonymous

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

94. anonymous

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

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. anonymous

yeah you're right @LogicalApple

97. anonymous

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

98. mathmate

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

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

100. mathmate

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

101. mathmate

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

102. anonymous

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

103. anonymous

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

104. mathmate

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

brb

106. anonymous

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. anonymous

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

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

109. anonymous

110. anonymous

The power set was generated by wolframalpha

111. mathmate

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

112. anonymous

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. anonymous

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

114. anonymous

Great job @LogicalApple Thank you!

115. anonymous

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

116. anonymous

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

117. anonymous

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

118. anonymous

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

119. anonymous

Yeah.

120. anonymous

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

121. anonymous

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

122. anonymous

It is actually closed haha.

123. anonymous

I mean, why so many responses still?

124. anonymous

Yeah because @mathmate still negates our answer hehe.

125. mathmate

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

126. mathmate

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

127. mathmate

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

@No-data What is your answer by the way, I already presented mine, with 30 elements.

129. anonymous

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

130. anonymous

Look a this its more clear.

131. mathmate

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. anonymous

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

133. anonymous

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

134. mathmate

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. anonymous

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

136. mathmate

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. anonymous

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

138. mathmate

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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