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A proof the Empty Set is unique

  1. Dec 31, 2014 #1
    1. The problem statement, all variables and given/known data
    The problem is to prove that there is only one empty set.

    Let A and B be empty sets,

    A is a subset of B and B is a subset of A (by the definition that the empty set is a subset of every set)

    So A=B (by definition)

    By convention, all empty sets are equal. Therefore, there is only one empty set.


    I'm fine with everything except the last line. I don't know if its my wording but something just doesn't feel right about it.
     
  2. jcsd
  3. Dec 31, 2014 #2
    It does make sense. The last line seems to be a form of the statement that "the Empty Set" is unique. The proof you presented shows that any two arbitrarily given empty sets are equal. For instance, if you were to say without loss of generality that set A is "the" empty set and set B is another arbitrarily given empty set, the fact that they are equal is proof that any set which is considered an empty set must be equivalent to "the" empty set.

    Does that help at all?
     
  4. Dec 31, 2014 #3
    That does. The more specific I can make it, the better. Thank you!
     
  5. Dec 31, 2014 #4
    One suggestion I will make is that you take out the phrase "by convention" if you can and make it more centered around what you mention in the proof. I could be wrong, but I have always understood the phrase "by convention" to imply some sort of already taken for granted assumption. If you cut that and say something pointing out the fact that since any set which has all of the properties of the Empty Set must be, by your proof, equivalent to the Empty Set, it keeps it from implying any general assumption.

    I of course could be off base with my understood meaning of "by convention", but still just a suggestion. :)
     
  6. Jan 1, 2015 #5

    haruspex

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    I would have thought that the definition of two sets being equal is that every element of one is also an element of the other. Uniqueness of the empty set follows fairly swiftly.
     
  7. Jan 1, 2015 #6

    HallsofIvy

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    Right. Suppose A and B are empty sets. What can we say about the statement "if x is in A then x is in B"? What about "if x is in B then x is in A"?
    (Note: if P is a false statement then "if P then Q" is a true statement for any Q.)

    Charles Stark, I think your proof is good until you suddenly state "By convention all empty sets are equal". If it were true by "convention" you couldn't prove it!
     
  8. Jan 1, 2015 #7

    Stephen Tashi

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    What definition of "empty set" are you using? If it is a theorem that the empty subset is a subset of every set then don't say that result is true "by the definition".
     
  9. Jan 3, 2015 #8
    Ooof Logic! I need to brush up on truth tables. As for the "convention", that's what was bothering me the most. I was reading some proofs on related topics and I saw a lot of "by convention" I think I just threw it on there.
     
  10. Jan 3, 2015 #9
    Another mistake, I thought it was listed as a definition in my book but it was listed as a convention. They asked for the class to reference definitions or statements from the book. Thank you!
     
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