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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

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  1. No-data Group Title
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    I think that the result is the empty set.

    • one year ago
  2. No-data Group Title
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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.

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

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

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

    • one year ago
  6. jim_thompson5910 Group Title
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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

    • one year ago
  7. No-data Group Title
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    That is what I did.

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

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

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

    • one year ago
  11. No-data Group Title
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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)

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

    • one year ago
  13. jim_thompson5910 Group Title
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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} {} }

    • one year ago
  14. jim_thompson5910 Group Title
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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} {} }

    • one year ago
  15. No-data Group Title
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    \(1\not \in S\) for example

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

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

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

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

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

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

    • one year ago
  22. jim_thompson5910 Group Title
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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)

    • one year ago
  23. jim_thompson5910 Group Title
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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} }

    • one year ago
  24. jim_thompson5910 Group Title
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    heres a page on the power set http://en.wikipedia.org/wiki/Power_set

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

    • one year ago
  26. jim_thompson5910 Group Title
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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)"

    • one year ago
  27. No-data Group Title
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    \(B-A=\{x\in B | x\not \in A\}\)

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

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

    • one year ago
  30. jim_thompson5910 Group Title
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    so P(S) - S would be start with P(S) and kick out any elements you find in S

    • one year ago
  31. No-data Group Title
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    Yeah, and there is no elements that are in both sets.

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

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

    • one year ago
  34. mathmate Group Title
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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).

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

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

    • one year ago
  37. mathmate Group Title
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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.

    • one year ago
  38. No-data Group Title
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    Yeah but why are you removing S?

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

    • one year ago
  40. No-data Group Title
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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.

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

    • one year ago
  42. jim_thompson5910 Group Title
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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

    • one year ago
  43. No-data Group Title
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    You can't remove 1 because this element is not in \(\wp S\)

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

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

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

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

    • one year ago
  48. jim_thompson5910 Group Title
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    yes i do

    • one year ago
  49. jim_thompson5910 Group Title
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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

    • one year ago
  50. LogicalApple Group Title
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    What if A = {1, {1}}, what would its power set look like?

    • one year ago
  51. LogicalApple Group Title
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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.

    • one year ago
  52. jim_thompson5910 Group Title
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    probably this, but I'm not 100% sure P(S) = { {1,{1}} {1}, {{1}} {} }

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

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

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

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

    • one year ago
  57. wio Group Title
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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 \(\wp S - S = \{\emptyset , \{1\}, \{2\}\}\)

    • one year ago
  58. swissgirl Group Title
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    I was thinking along the same lines

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

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

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

    • one year ago
  62. wio Group Title
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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.

    • one year ago
  63. wio Group Title
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    Except perhaps the empty set?

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

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

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

    • one year ago
  67. No-data Group Title
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    mmm ok.

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

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

    • one year ago
  70. mathmate Group Title
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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.

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

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

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

    • one year ago
  74. mathmate Group Title
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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).

    • one year ago
  75. No-data Group Title
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    I agree with \(S\subseteq\wp S\).

    • one year ago
  76. mathmate Group Title
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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.

    • one year ago
  77. No-data Group Title
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    Ok. now list all the elements of S.

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

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

    • one year ago
  80. mathmate Group Title
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    brb

    • one year ago
  81. No-data Group Title
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    ok

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

    • one year ago
  83. mathmate Group Title
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    * 31 elements.

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

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

    • one year ago
  86. mathmate Group Title
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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) )

    • one year ago
  87. No-data Group Title
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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.

    • one year ago
  88. No-data Group Title
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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.

    • one year ago
  89. mathmate Group Title
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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).

    • one year ago
  90. No-data Group Title
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    yeas. the power set of S has set elements. then you can't remove non set element from it.

    • one year ago
  91. LogicalApple Group Title
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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?

    • one year ago
  92. LogicalApple Group Title
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    Is 1 a subset of {1, 2} ?

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

    • one year ago
  94. LogicalApple Group Title
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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.

    • one year ago
  95. mathmate Group Title
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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.

    • one year ago
  96. No-data Group Title
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    yeah you're right @LogicalApple

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

    • one year ago
  98. mathmate Group Title
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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.

    • one year ago
  99. mathmate Group Title
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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.

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

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

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

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

    • one year ago
  104. mathmate Group Title
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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} \( \notin S \)

    • one year ago
  105. mathmate Group Title
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    brb

    • one year ago
  106. LogicalApple Group Title
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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.

    • one year ago
  107. No-data Group Title
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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

    • one year ago
  108. mathmate Group Title
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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.

    • one year ago
  109. LogicalApple Group Title
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    • one year ago
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  110. LogicalApple Group Title
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    The power set was generated by wolframalpha

    • one year ago
  111. mathmate Group Title
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    Note that 1 is an element of S, but {1} is not, it is a subset of S.

    • one year ago
  112. LogicalApple Group Title
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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.

    • one year ago
  113. LogicalApple Group Title
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    *P also has {}, but S does not.

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

    • one year ago
  115. No-data Group Title
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    by the way the empty set is a subset of S but is not an element of S @mathmate

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

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

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

    • one year ago
  119. wio Group Title
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    Yeah.

    • one year ago
  120. LogicalApple Group Title
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    I would think that expression holds unless S contains an element that is also a subset of S.

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

    • one year ago
  122. No-data Group Title
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    It is actually closed haha.

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

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

    • one year ago
  125. mathmate Group Title
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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.

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

    • one year ago
  127. mathmate Group Title
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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.

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

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

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

    • one year ago
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  131. mathmate Group Title
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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} }

    • one year ago
  132. No-data Group Title
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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.

    • one year ago
  133. No-data Group Title
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    For example \( \emptyset \subset S\) but \(\emptyset \not \in S \)

    • one year ago
  134. mathmate Group Title
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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.

    • one year ago
  135. No-data Group Title
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    Then it's not correct to remove the empty set from the powerset in this case right?

    • one year ago
  136. mathmate Group Title
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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.

    • one year ago
  137. No-data Group Title
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    No the empty set is not in every set. The empty set is a subset of every set.

    • one year ago
  138. mathmate Group Title
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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.

    • one year ago
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