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No-data

  • one year 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
    • one year ago
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    I think that the result is the empty set.

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

  4. No-data
    • one year ago
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    Am I right?

  5. No-data
    • one year ago
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    @brinethery what do you think?

  6. jim_thompson5910
    • one year 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
    • one year ago
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    That is what I did.

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

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

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

  11. No-data
    • one year 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
    • one year ago
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    and none of the elements of S are in P(S)

  13. jim_thompson5910
    • one year 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
    • one year 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
    • one year ago
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    \(1\not \in S\) for example

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

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

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

  19. jim_thompson5910
    • one year ago
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    not quite, P(S) - S does not equal P(S)

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

  21. No-data
    • one year 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
    • one year 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
    • one year 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
    • one year ago
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    heres a page on the power set http://en.wikipedia.org/wiki/Power_set

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

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

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

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

  30. jim_thompson5910
    • one year 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
    • one year ago
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    Yeah, and there is no elements that are in both sets.

  32. jim_thompson5910
    • one year ago
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    so then your original answer of empty set works

  33. No-data
    • one year 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
    • one year 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
    • one year ago
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    no @mathmate you're mixing sets with elements too.

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

  37. mathmate
    • one year 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
    • one year ago
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    Yeah but why are you removing S?

  39. No-data
    • one year 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
    • one year 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
    • one year ago
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    So you're removing nothing from \(\wp S\)

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

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

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

  46. No-data
    • one year ago
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    lol no problem.

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

  48. jim_thompson5910
    • one year ago
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    yes i do

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

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

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

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

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

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

  57. wio
    • one year 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
    • one year ago
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    I was thinking along the same lines

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

  60. No-data
    • one year ago
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    This problem is tricky

  61. wio
    • one year ago
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    Ohhh, I read it wrong then.

  62. wio
    • one year 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
    • one year ago
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    Except perhaps the empty set?

  64. No-data
    • one year ago
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    yeah exactly !

  65. No-data
    • one year ago
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    not, the empty set is not in S

  66. wio
    • one year 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
    • one year ago
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    mmm ok.

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

  69. No-data
    • one year 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
    • one year 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
    • one year ago
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    But you have no reason to remove the last item @mathmate

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

  73. No-data
    • one year ago
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    Because \(S\not \in S\)

  74. mathmate
    • one year 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
    • one year ago
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    I agree with \(S\subseteq\wp S\).

  76. mathmate
    • one year 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
    • one year ago
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    Ok. now list all the elements of S.

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

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

  80. mathmate
    • one year ago
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    brb

  81. No-data
    • one year ago
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    ok

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

  83. mathmate
    • one year ago
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    * 31 elements.

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

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

  86. mathmate
    • one year 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
    • one year 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
    • one year 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
    • one year 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
    • one year 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
    • one year 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
    • one year ago
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    Is 1 a subset of {1, 2} ?

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

  94. LogicalApple
    • one year 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
    • one year 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
    • one year ago
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    yeah you're right @LogicalApple

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

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

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

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

  104. mathmate
    • one year 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
    • one year ago
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    brb

  106. LogicalApple
    • one year 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
    • one year 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
    • one year 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
    • one year ago
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  110. LogicalApple
    • one year ago
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    The power set was generated by wolframalpha

  111. mathmate
    • one year 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
    • one year 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
    • one year ago
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    *P also has {}, but S does not.

  114. No-data
    • one year ago
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    Great job @LogicalApple Thank you!

  115. No-data
    • one year 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
    • one year ago
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    I don't understand what the problem here is... \[ \mathcal P(S) - S = \mathcal P(s) \]

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

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

  119. wio
    • one year ago
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    Yeah.

  120. LogicalApple
    • one year 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
    • one year 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
    • one year ago
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    It is actually closed haha.

  123. wio
    • one year ago
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    I mean, why so many responses still?

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

  125. mathmate
    • one year 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
    • one year ago
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    Therefore, by definition of \ , {} gets kicked out of P(S)-S.

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

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

  130. No-data
    • one year ago
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    Look a this its more clear.

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  131. mathmate
    • one year 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
    • one year 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
    • one year ago
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    For example \( \emptyset \subset S\) but \(\emptyset \not \in S \)

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

  136. mathmate
    • one year 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
    • one year 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
    • one year 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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