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No-data

  • 2 years ago

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

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  1. No-data
    • 2 years ago
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    I think that the result is the empty set.

  2. No-data
    • 2 years ago
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    \[\wp S - S = \emptyset\] because there is not an element on S that equals an elemento of the power set of S.

  3. No-data
    • 2 years ago
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    \(\not \exists x\in S\;|\; x\in\wp S \)

  4. No-data
    • 2 years ago
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    Am I right?

  5. No-data
    • 2 years ago
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    @brinethery what do you think?

  6. jim_thompson5910
    • 2 years ago
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    If \[\wp S\] is the power set, then list out everything in the power set but kick out the set S

  7. No-data
    • 2 years ago
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    That is what I did.

  8. jim_thompson5910
    • 2 years ago
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    ok, but you won't get the empty set

  9. No-data
    • 2 years ago
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    But I didn't kick out the set S.

  10. jim_thompson5910
    • 2 years ago
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    that's the what the " - S " portion is saying

  11. No-data
    • 2 years ago
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    What I think it says is that I must kick out the elements of S that are elements of P(S)

  12. No-data
    • 2 years ago
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    and none of the elements of S are in P(S)

  13. jim_thompson5910
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    \(1\not \in S\) for example

  16. No-data
    • 2 years ago
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    because \(1\not = \{1\}\)

  17. No-data
    • 2 years ago
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    And the same happens with 2,3,4,5

  18. No-data
    • 2 years ago
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    Mmm then I think \(\wp S - S = \wp S\)

  19. jim_thompson5910
    • 2 years ago
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    not quite, P(S) - S does not equal P(S)

  20. jim_thompson5910
    • 2 years ago
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    close though, because you're only erasing one element of the powerset

  21. No-data
    • 2 years ago
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    I cant quit {1,2,3,4,5} because \(\{1,2,3,4,5\}\not \in S\)

  22. jim_thompson5910
    • 2 years ago
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    {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
    • 2 years ago
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    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
    • 2 years ago
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    heres a page on the power set http://en.wikipedia.org/wiki/Power_set

  25. No-data
    • 2 years ago
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    I'm agree with the power set expansion. But I think you're difference is wrong

  26. jim_thompson5910
    • 2 years ago
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    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
    • 2 years ago
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    \(B-A=\{x\in B | x\not \in A\}\)

  28. No-data
    • 2 years ago
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    What are you saying is this: \(\wp S - \{S\} \)

  29. jim_thompson5910
    • 2 years ago
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    oh good point, I'm mixing up sets and elements here...

  30. jim_thompson5910
    • 2 years ago
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    so P(S) - S would be start with P(S) and kick out any elements you find in S

  31. No-data
    • 2 years ago
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    Yeah, and there is no elements that are in both sets.

  32. jim_thompson5910
    • 2 years ago
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    so then your original answer of empty set works

  33. No-data
    • 2 years ago
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    No, I think the right answer is \(\wp S - S = \wp S\) because there is no elements to quit from \(\wp S\)

  34. mathmate
    • 2 years ago
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    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
    • 2 years ago
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    no @mathmate you're mixing sets with elements too.

  36. No-data
    • 2 years ago
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    you can't quit \(S\) because \(S\not \in S\)

  37. mathmate
    • 2 years ago
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    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
    • 2 years ago
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    Yeah but why are you removing S?

  39. No-data
    • 2 years ago
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    When you do a set difference you remove the elements that both sets have in common right?

  40. No-data
    • 2 years ago
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    You must remove all the elements that are in \(\wp S \) and in \(S\) but I can't find those elements.

  41. No-data
    • 2 years ago
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    So you're removing nothing from \(\wp S\)

  42. jim_thompson5910
    • 2 years ago
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    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
    • 2 years ago
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    You can't remove 1 because this element is not in \(\wp S\)

  44. No-data
    • 2 years ago
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    \(1\not = \{1\}\) for example

  45. jim_thompson5910
    • 2 years ago
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    true, not thinking on this one lol

  46. No-data
    • 2 years ago
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    lol no problem.

  47. No-data
    • 2 years ago
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    So do you agree with my answer @jim_thompson5910 ?\[\wp S - S = \wp S\]

  48. jim_thompson5910
    • 2 years ago
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    yes i do

  49. jim_thompson5910
    • 2 years ago
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    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
    • 2 years ago
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    What if A = {1, {1}}, what would its power set look like?

  51. LogicalApple
    • 2 years ago
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    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
    • 2 years ago
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    probably this, but I'm not 100% sure P(S) = { {1,{1}} {1}, {{1}} {} }

  53. No-data
    • 2 years ago
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    \[\wp A = \{\emptyset, \{1\}, \{\{1\}\}, A\}\]

  54. LogicalApple
    • 2 years ago
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    In that instance, the original set contains the element {1}, but so does the power set.

  55. No-data
    • 2 years ago
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    Yes you're right @LogicalApple that relation holds for sets with non-set elements.

  56. No-data
    • 2 years ago
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    I mean \(\wp A - A \not = \wp A\)

  57. wio
    • 2 years ago
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    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
    • 2 years ago
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    I was thinking along the same lines

  59. No-data
    • 2 years ago
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    You can take out the set itself if you have the following expression: \[\wp S - \{S\}\]

  60. No-data
    • 2 years ago
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    This problem is tricky

  61. wio
    • 2 years ago
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    Ohhh, I read it wrong then.

  62. wio
    • 2 years ago
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    So basically, it's just the power set then, because the power set does NOT have any elements of S in it.

  63. wio
    • 2 years ago
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    Except perhaps the empty set?

  64. No-data
    • 2 years ago
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    yeah exactly !

  65. No-data
    • 2 years ago
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    not, the empty set is not in S

  66. wio
    • 2 years ago
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    Yeah, in this case it there isn't any elements in common, but I was just thinking generally.

  67. No-data
    • 2 years ago
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    mmm ok.

  68. wio
    • 2 years ago
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    I'm used to seeing the power set denoted with \mathcal P : \(\mathcal P(S) \)

  69. No-data
    • 2 years ago
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    Yeah there are several way of writting it. I just use the same as the one in my book hehe.

  70. mathmate
    • 2 years ago
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    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
    • 2 years ago
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    But you have no reason to remove the last item @mathmate

  72. mathmate
    • 2 years ago
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    Not at all, but that's what the question wants.

  73. No-data
    • 2 years ago
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    Because \(S\not \in S\)

  74. mathmate
    • 2 years ago
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    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
    • 2 years ago
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    I agree with \(S\subseteq\wp S\).

  76. mathmate
    • 2 years ago
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    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
    • 2 years ago
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    Ok. now list all the elements of S.

  78. No-data
    • 2 years ago
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    And quit the elements that are in S and the power set of S

  79. mathmate
    • 2 years ago
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    Note that P(S) does not have any element like 1,2,3,... , only subsets.

  80. mathmate
    • 2 years ago
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    brb

  81. No-data
    • 2 years ago
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    ok

  82. mathmate
    • 2 years ago
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    So P(S)\S will be left with only 31 subsets, the first 31.

  83. mathmate
    • 2 years ago
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    * 31 elements.

  84. No-data
    • 2 years ago
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    Why you're removing the set S?

  85. No-data
    • 2 years ago
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    S is not an element of S, so you can't remove it.

  86. mathmate
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    You must remove from the power set of S the elements of S that are in the power set of S too.

  89. mathmate
    • 2 years ago
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    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
    • 2 years ago
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    yeas. the power set of S has set elements. then you can't remove non set element from it.

  91. LogicalApple
    • 2 years ago
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    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
    • 2 years ago
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    Is 1 a subset of {1, 2} ?

  93. No-data
    • 2 years ago
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    1 is not a set. so it can't be a subset.

  94. LogicalApple
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    yeah you're right @LogicalApple

  97. No-data
    • 2 years ago
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    So you agree with this @mathmate \[\wp S - S = \wp S\]

  98. mathmate
    • 2 years ago
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    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
    • 2 years ago
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    Sorry, I have a different opinion. P(S) has 32 elements (which are subsets), while P(S)-S has only 31.

  100. mathmate
    • 2 years ago
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    I meant to say in the previous post: "By definition, the power set cannot contain elements of the original set".

  101. mathmate
    • 2 years ago
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    In fact, in general, P(S)-S has 2^n-1 elements.

  102. No-data
    • 2 years ago
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    Remember the definition of a set difference: \[A-B=\{x\in A|x\not \in B\}\]

  103. No-data
    • 2 years ago
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    I'm not looking for the number of elements. but for a resulting set.

  104. mathmate
    • 2 years ago
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    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
    • 2 years ago
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    brb

  106. LogicalApple
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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  110. LogicalApple
    • 2 years ago
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    The power set was generated by wolframalpha

  111. mathmate
    • 2 years ago
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    Note that 1 is an element of S, but {1} is not, it is a subset of S.

  112. LogicalApple
    • 2 years ago
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    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
    • 2 years ago
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    *P also has {}, but S does not.

  114. No-data
    • 2 years ago
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    Great job @LogicalApple Thank you!

  115. No-data
    • 2 years ago
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    by the way the empty set is a subset of S but is not an element of S @mathmate

  116. wio
    • 2 years ago
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    I don't understand what the problem here is... \[ \mathcal P(S) - S = \mathcal P(s) \]

  117. LogicalApple
    • 2 years ago
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    @wio I have a tough problem ready for you later though..

  118. No-data
    • 2 years ago
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    that expression hold for \(S = \{1,2,3,4,5\}\)

  119. wio
    • 2 years ago
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    Yeah.

  120. LogicalApple
    • 2 years ago
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    I would think that expression holds unless S contains an element that is also a subset of S.

  121. wio
    • 2 years ago
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    It holds when: \[ \neg \exists A \subseteq S \wedge A \in S \] Why is this question not closed?

  122. No-data
    • 2 years ago
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    It is actually closed haha.

  123. wio
    • 2 years ago
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    I mean, why so many responses still?

  124. No-data
    • 2 years ago
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    Yeah because @mathmate still negates our answer hehe.

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

  126. mathmate
    • 2 years ago
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    Therefore, by definition of \ , {} gets kicked out of P(S)-S.

  127. mathmate
    • 2 years ago
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    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
    • 2 years ago
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    @No-data What is your answer by the way, I already presented mine, with 30 elements.

  129. No-data
    • 2 years ago
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    The empty set can't be removed because is not an element of S neither S.

  130. No-data
    • 2 years ago
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    Look a this its more clear.

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  131. mathmate
    • 2 years ago
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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} }

  132. No-data
    • 2 years ago
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    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
    • 2 years ago
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    For example \( \emptyset \subset S\) but \(\emptyset \not \in S \)

  134. mathmate
    • 2 years ago
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    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
    • 2 years ago
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    Then it's not correct to remove the empty set from the powerset in this case right?

  136. mathmate
    • 2 years ago
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    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
    • 2 years ago
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    No the empty set is not in every set. The empty set is a subset of every set.

  138. mathmate
    • 2 years ago
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    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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