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Proof that empty set is subset of every set

  1. Sep 30, 2012 #1
    My book presents the theorem that the empty set is a subset of every set as follows:

    Let A be a set.
    Let x be any object.
    Assume x belongs to empty set.
    Then x belongs to A.
    Thus the empty set is a subset of A.


    In other words, the proposition [(x belongs to empty set) implies (x belongs to A)] is true because the antecedent (x belongs to empty set) is always false, and a false statement may imply anything. But I really don't like this proof and I think it contradicts what my book previously stated: "We must be careful about making assumptions, because we can only be certain that what we proved will be true when all the assumptions are true."
    But clearly the assumption that x belongs to empty set is always false.

    I mean if you take this to be a valid proof, then I could just as well say that
    [(x belongs to empty set) implies (empty set is not subset of A)]
    thereby proving that the empty set is not a subset of any set...

    So my question is, is a proof absolutely correct even if it is based on false assumptions?

    I can see how to prove this theorem by contradiction: if it is not the case that [(x belongs to empty set) implies (x belongs to A)], then
    [(x belongs to empty set) and (x doesnt belong to A)] is true.
    But x doesn't belong to empty set. So the above is false, which is contradiction, and so empty set is subset of A.

    But i'm just not convinced that the book's proof is valid...
     
  2. jcsd
  3. Sep 30, 2012 #2
    I don't get it either.
     
  4. Sep 30, 2012 #3

    chiro

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    Science Advisor

    Hey Aziza.

    I don't know about that proof (looks kinda dodgy), but for the purposes of the proof I'd go with showing empty_set Intersection Set = empty_set for any set.

    This is the basic idea of a subset that if something is a subset of something else (like A is a subset of B) then A Intersection B = A for all appropriate A and B. If something is not a subset, then you'll get a result that "trims" A and gives you a "partial" version of A (something that has been truncated).
     
  5. Sep 30, 2012 #4

    Stephen Tashi

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    The "if..." part of "if x is an element of the empty set" is not the "if..." part of what you are proving.

    To translate "The empty set is a subset of every set" in to an "if....then...." statement, you must say something like "if E is the empty set and A is a set then E is a subset of A".

    (If you want to be even more precise, you have to use the quantifier "for each" and say "If E is the empty set then for each set A, E is a subset of A".)

    The proof only assumes "E is the empty set" is true. It doesn't begin by assuming "there is some element x that is a subset of the empty set E". So the proof given by the book does not begin by assuming something that cannot be true.

    The fact that a false statement implies any statement is disturbing to many people. But consider the alternatives. Suppose you say "If x i> 0 0 then 1/x > 0 " and someone says "No! You're wrong. Suppose x = -13. Then 1/(-13) isn't greater than 0". We don't regard the example of x = -13 as a disproof. The statement "if x > 0 then 1/x > 0" remains true when x = -13.

    There might be some who want to invent a third classification for statements so each statement must be one of: { True, False, Undecided}. However, the traditional way of using only (True, False} works for mathematics. We don't say that "if x > 0 then 1/x > 0" is "Undecided" when x = -13. We say the statement is True. ( One way to look at it is that we are saying the "if ..." part correctly filters out all situations that aren't relevant to the context of the conclusion.)
     
    Last edited: Sep 30, 2012
  6. Sep 30, 2012 #5
    This is a special case of the fact that a false proposition implies a true one. For example if 2+2=5 then I am the Pope. If you believe that we're good; otherwise ask and we can talk about it more.

    If you understand that False implies True, then let x be an element of the empty set. That's already impossible. If x is a member of the empty set then x is a purple elephant. Therefore the empty set is a subset of the set of purple elephants.

    This is counterintuitive the first time you hear it; but at some point it will begin to seem natural. Just remember that False implies True and that if x is in the empty set then x is a tuna sandwich. Isn't that actually true?
     
  7. Oct 1, 2012 #6

    MarneMath

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    The way I like to go about it is like this:

    Let S be a subset of T. Every x in S is also in T. Thus there does not exist an element in S that does not exist T.

    We also say that there are no elements of the empty set by definition. Thus the empty set also has no elements that are also not in any other set. So vacuously, we say all elements of the empty sets are in every other set.

    The way your book did it seems dodgy at best and as everyone has pointed out can basically prove anything else you want to prove.
     
  8. Oct 1, 2012 #7

    Stephen Tashi

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    I disagree. Aziza's description of the proof is not a complete proof, but it is an outline of the standard way of proving this theorem. It does not outline a procedure that would let you prove "anything else you want to prove".
     
  9. Oct 1, 2012 #8
    I don't like the proof either, particularly the statement "let x be an element of the empty set". It is more neatly stated as if the empty set is a subset of A then every member of the empty set is a member of A. But the empty set has no members so the condition is automatically fulfilled.

    I prefer the proof by contradiction. Assume that the empty set is not a subset of A. Then there exists an element of the empty set which is not an element of A. But the empty set contains no elements (and thus no elements which are not in A). Therefore, the empty set is a subset of A. Much nicer without the bothersome statement.
     
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