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The empty set

  1. Aug 9, 2004 #1
    When I took a math foundations class we only did naive set theory and took as an axiom that the empty set is a member of every set. The book had formal set theory and thus listed the ZFC axioms. One of them was that the empty set exists and that it was a member of every set. I've looked at a couple of listing of the axioms on the net and they only give the existence of the empty set.

    To prepare for graduate analysis (I start grad school in a week), I've been reading through Rudin's Principles of Mathematical Analysis. I noticed today that an exercise in the book is to prove that the empty set is a member of every set.

    I'm not asking for the solution, but I was wondering how one would go about proving this. Whenever you prove subsets, you chase elements. But the empty set has no elements. Are there other methods of proving subsets?
  2. jcsd
  3. Aug 9, 2004 #2
    Do you mean "prove that the empty set is a subset of every set"?
  4. Aug 9, 2004 #3
    Yeah, sorry about that. I did, of course, mean subset, not member.
  5. Aug 9, 2004 #4
    Well, any statement of the form "if [itex]x\in\varnothing[/itex] then ..." is always a true statement, no matter what "..." is, since the condition is always false.
  6. Aug 10, 2004 #5
    I think I figured it out. By definition, A is a subset of B provided that if a is in A, then a is in B.

    Let X be non-empty set and assume that {} is not a subset of X. Then there exists an x in {} such that x is not in X. A contradiction, for all x, x is not in {}. Therefore {} is a subset of X. qed
  7. Aug 10, 2004 #6
    That's also a good approach. Take advantage of the fact that [tex]\exists x\in\varnothing[/tex] is always false.
    Last edited: Aug 10, 2004
  8. Aug 11, 2004 #7


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    It seems to me that this postulate is a notion due to thinking about abstractions and not what the sets stand for in the world outside the mind. It seems to me that a set X i. e. something does not necessarily include Empty set i. e. nothing.
    I. e., Something does not always contain nothing as part of something. Just a theory.
  9. Aug 11, 2004 #8
    All that's taken as an axiom is that the empty set exists. I provided a proof above that the empty set is a subset of every non-empty set. Also, set are only abstractions. I don't know what you mean by "in the world outside the mind" wrt sets.
  10. Aug 11, 2004 #9
    As stated earlier, the empty set is a subset of every set because the conditional IF/THEN is always true when the antecedent (the part after the IF) is false. This is known as being vacuously true. So when I say if x is in the empty set then x is in the set A, the whole conditional is always (vacuously) true. In other words, the truth table for the conditional is T when you have F/T or F/F for the conditional IF/THEN. If you want to know WHY this is so, I can post an article on vacuous truth or you can search wikipedia.com for it...
  11. Aug 11, 2004 #10
    Thanks for the offer, but I learned all this way back in my first logic class. What I was asking for was a proof. And, as my analysis teacher would say, you have the elements of a proof in what you've written. From a formal perspective, I think what you've written and my proof and logically equivalent. But, in math, a 3 sentence proof is better than an equivalent paragraph.
  12. Aug 11, 2004 #11
    Well, you did ask for a method of proof and not an actual proof. When I gave you the vacuously true statement, I was trying to push you towards:

    [tex]x\in\varnothing\Rightarrow x\in X[/tex] is true therefore by definition [tex]\varnothing\subseteq X[/itex].
  13. Aug 11, 2004 #12
    You're exactly right. Maybe I read condecension in the last post where none existed...if so, I apologize.

    BTW, is there a link that explains how to use those latex symbols here?
  14. Aug 11, 2004 #13
    There's a thread in the General Physics forum that has a lot of info and links.

  15. Aug 26, 2006 #14


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    Singleton set actually has two elements.

    HiHo! :smile:

    I was reading the first chapter of a book titled Elements Of The Theory Of Computation 2/E when suddenly I thought that every set actually has one more member called "nothing" because an empty set is a subset of every set. For example A={1,2} is actually A={1,,2}. Hence, I came to a conclusion that singleton is no other than the empty set itself (i.e., C={}) because a singleton like C={3} actually has two members, one of which is "nothing" (i.e., C={3,}).

    Now I know the truth after reading this thread. But, I still wonder if I write a notation like this, A={1,,2}, is not that "nothing" represented by ,, ? If not, why not? If yes, well, I think my reasoning above is justifiable :biggrin:

  16. Aug 26, 2006 #15

    matt grime

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    You are perfectly at liberty to define this notation, and the corresponding theory. It will bear no relation to set theory as done by anyone else. And you'd at least need to check that it wasn't obviously inconsistent.
  17. Aug 26, 2006 #16


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    But the proof is not complete! This only proves that {} is a subset of every non-empty set.:rofl:

    Obviously, any set is a subset of itself by you have to add an extra line for that. That's why I think that appealing to the fact that "if x is a member of {}" is always false is a better proof.
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