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Set Theory : Discrete Mathematics

  1. Jul 14, 2012 #1
    1. The problem statement, all variables and given/known data

    The Question data is as follows :

    Consider the following sets, where U represents a universal set :

    U = {1, 2, 3, 4, ∅, {1}}
    A = {1, 3}
    B = {{1}, 1}
    C = {2 , 4}
    D = { ∅ , 1, 2 )


    2. Relevant equations

    Which one of the following statements is true ?

    1. The cardinality of U, ie |U| = |A| + |B| + |C| + |D|
    2. {} ⊆ U
    3. {∅, {1}} ∈ U
    4. U union B = B



    3. The attempt at a solution

    1. So on this one I attempted, it is clear that this is not the case, and this statement is false, due to the fact that the cardinality indicates the amount of elements included in the set. The set for U , and the other sets' cardinality differ. If you would like me to illustrate what I have found, let me know , I don't mind adding it here.

    2. This to my knowledge is true. If {} is an empty set, it does not have any element that is not in U, and is thus possible to 'reside' as a subset, in U. To my knowledge this is then true.

    3. To my knowledge, this one is also true. The problem I have with this question , is that only one can be true, so either 2 or 3 is true. A set can be an element of another set, like in this case ( I read up on a couple of other articles on this website - if I am wrong with this one, please point it out to me, so that I can assess my understanding of this subject).

    4. The union of U and B, will be the composition of all the elements in U, as well as the elements in B, without any duplication of elements. In a Venn diagram this will thus be the 2 circles, together, with the whole field ( of both circles) coloured in. So the sets together will be more than just the set B, and thus be incorrect.




    There is no further information required.


    Background from me : I was a LAW student, and changed to BSc Computer Science, and I am studying through Unisa in South Africa. This is part of one of my assignments for a subject COS1501, and the assignment is due next month, so I am not trying to catch up time by submitting my inquiry to you to answer, this is well in advance and my understanding is lacking , which I require assistance with.

    Your advise, even if you point me in the right direction, will be highly appreciated.
     
  2. jcsd
  3. Jul 14, 2012 #2

    Dick

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    I think you are ok on all of those. And I think more than one of them is correct.
     
  4. Jul 14, 2012 #3

    micromass

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    (3) is false. While it is true that [itex]\{1,\{1\}\}[/itex] is a subset of U, it is not an element of U.
     
  5. Jul 14, 2012 #4

    Dick

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    Dang it. I'm so sloppy these days. Thanks, micromass.
     
  6. Jul 14, 2012 #5
    Dick , Micromass, you guys are awesome!

    So, if I understand correctly, a set cannot be referred to as an element of a full set? I am using the term a 'full set' due to lack of a better word for 'U' - the set with all the elements.

    So :

    A = {1,2,3}
    a) this set A, has elements , 1 , and 2 , and 3.

    B = {1, 2, {3,4}}
    a) This set B, has element 1 , and element 2, however the 'element {3,4} is not referred to as an element, even though it is in the set, and referred to as a sub set of B, so in actual fact to make it practical it can be written :


    B = {1, 2}
    {3, 4) ⊆ B

    And this means that there is a sub set, and not an Element. This will also then be a proper subset, since not all the elements are present. Due to a subset being either a copy of the set, or alternatively a proper subset, it will then be correct to write it either way, as ⊆ or as '⊆ without the bottom line'.


    If my understanding of the above is correct, please let me know, then I am happy.


    Will it then be possible that anything inside a curly bracket / brace, can be referred to as an element ? To me , if the elements are in a curly bracket / brace, it cannot be referred to as an element.

    {} cannot be ∈ of A
    {} can be the set, or subset, of a set A.

    Thanks again guys, I dont know what I should be adding in the discussion to make it as easy as possible to understand, however if you find that I am not adding something , please let me know, and I will add it. ( It is 05:30 AM in South Africa, so I am going to take a quick nap).
     
  7. Jul 15, 2012 #6

    HallsofIvy

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    Yes, that is correct.

    No, this is wrong. B has three elements, 1, 2, and {3, 4}. {3, 4} definitely is an element of B, NOT a subset. {1, 2} would be a subset of B because its elements, 1 and 2 are also elements of B. The elements of {3, 4}, 3 and 4, are NOT elements of B.

    Again this is wrong. You are told to begin with that B= {1, 2, {3, 4}} so it cannot be equal to {1, 2}. And, if B were defined as {1, 2} then it would NOT be the case that {3,4} ⊆ B because, in this case, 3 and 4 are NOT elements of B.

    Once more, no. You are making this more complicated than it should be. The notation
    "B= {1, 2, {3, 4}}" means exactly what it says- B has three elements, 1, 2, and the set {3, 4}.


    Sorry, but it is not correct.


    Everything inside "curly braces" is an element of a set. Sets themselves can be a member of a set.

    {} can be an element of a set, if you define the set that way. In fact, your original post you defined U = {1, 2, 3, 4, ∅, {1}}. ∅ is just another notation for {}, the empty set and certainly is an element of this set. It also happens to be a subset- the empty set is a subset of any set.

    Unfortunately, most of what you think you know here is incorrect. I recommend you review "element' and "subset".
     
  8. Jul 15, 2012 #7
    HallsofIvy -I just want to say that I appreciate your feedback, so so much. I will be looking into this feedback.

    This makes sense, and the proof is in the pudding , so I will be doing a couple of exercises again to familiarise me with this work.

    Thanks again!
     
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