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Two Empty Sets

  1. Nov 6, 2012 #1
    I want to ask that, are you empty sets equal ????
     
  2. jcsd
  3. Nov 6, 2012 #2

    haruspex

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    I reply, therefore I am not an empty set.
    Yes, two empty sets are equal, even if you regard them as subsets of entirely different beasts. So any two power sets have a nonempty intersection.
     
  4. Nov 6, 2012 #3
    How can two Empty Sets be equal ???
    take a look at this example..
    1st set: Number of trees in the forest
    2nd Set: Number of Student of in the class

    suppose that these are empty sets.. how are they equal ??? they both have different description..
     
  5. Nov 6, 2012 #4

    haruspex

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    You've defined two numbers, not two sets, but passing over that...
    "A whole number between 1 and 3." "Positive square root of 4." Two descriptions, equal answers.
     
  6. Nov 6, 2012 #5

    Erland

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    Two sets are equal if every member of the first set is also a member of the second set, and vice versa. If two empty set were not equal, there would be one element in one of the sets which is not a member of the other set, and this is impossible, since an empty set has no member at all, whether member or not member of another empty set.
     
  7. Nov 6, 2012 #6

    Bacle2

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    If two empty sets were different, one of them would contain an element

    not contained in the other....
     
  8. Nov 7, 2012 #7

    HallsofIvy

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    Just to put my oar in: two sets, A and B, are equal if and only both statements
    "if x is in A then x is in B" and "if x is in B then x is in A" are true.

    If A is the empty set, then "if x is in A" is false so the statement "if x is in A then x is in B" is trivially true. If B is the empty set then "if x is in A" is false so the statement "if x is in B then x is in A" is trivially true. Therefore, if A and B are both empty, then they are equal.
     
  9. Nov 7, 2012 #8
    The empty set is unique and the following proof ascertains that.

    Suppose that there is another empty set denoted by ##\emptyset'## ,then we have:

    1)##\forall A[A\cup\emptyset =A]##

    2)##\forall A[A\cup\emptyset' =A]##

    In (1) we put ##A =\emptyset'## and we get: ##\emptyset'\cup\emptyset =\emptyset'##

    In (2) we put ## A =\emptyset## and we get :##\emptyset\cup\emptyset' =\emptyset##

    But since ,##\emptyset'\cup\emptyset = \emptyset\cup\emptyset' ##

    We can conclude :##\emptyset'=\emptyset ##

    Hence all the empty sets are equal
     
    Last edited: Nov 7, 2012
  10. Nov 7, 2012 #9
    Here is another proof that the empty set is unique:

    Suppose again that there is another empty set denoted by,##\emptyset'##

    Then we have:

    1) ##\forall A[\emptyset\subseteq A]##

    2)##\forall A[\emptyset'\subseteq A]##

    In (1) put : ##A=\emptyset'## and we get : ##\emptyset\subseteq \emptyset']##.....................................................................3

    In (2) put : ##A=\emptyset## and we get : ##\emptyset'\subseteq \emptyset]##.....................................................................4

    And from (3) and ( 4) we conclude :,##\emptyset' = \emptyset##
     
  11. Nov 7, 2012 #10

    Bacle2

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    I don't get it: you showed ##A\cup\ B=B\cup\ A## for A,B being (supposedly)

    different copies of the empty set. But

    this is true for any two sets, since union is commutative; I don't see how

    this shows that A=B.
     
  12. Nov 8, 2012 #11
    WE have :

    ##A\cup B= B\cup A## for all A,B

    Then put :##A=\emptyset ## and ## B=\emptyset'## and we have:

    ##\emptyset\cup\emptyset' = \emptyset'\cup\emptyset ##

    But ,##\emptyset\cup\emptyset'=\emptyset ## and ##\emptyset'\cup\emptyset =\emptyset'## as i have shown in my previous proof

    Hence ##\emptyset'= \emptyset ##
     
  13. Nov 8, 2012 #12
    Those proofs strike me as a bit convoluted. The fact that all empty sets are equal is a consequence of the axiom of extensionality, which is literally the most fundamental property of sets, which says that sets are equal if and only if they have the same members. If the set A has no members and the set B has no members, then by extensionality A = B.
     
  14. Nov 8, 2012 #13
    This is wrong as the followinf argument shows:

    WE know that:

    ##A=B\Longleftrightarrow [(A\subseteq B)\wedge(B\subseteq A)##.

    Hence:

    ##A\neq B\Longleftrightarrow[\neg(A\subseteq B)\vee \neg(B\subseteq A)]##.

    OR

    ##A\neq B\Longleftrightarrow[\exists x(x\in A\wedge \neg x\in B)]\vee[\exists x(x\in B\wedge \neg x\in A)]##

    Now if we put : ##A =\emptyset## and ##B=\emptyset'## ,we get that:

    ##\emptyset\neq \emptyset'\Longleftrightarrow[\exists x(x\in \emptyset\wedge \neg x\in \emptyset')]\vee[\exists x(x\in \emptyset'\wedge \neg x\in \emptyset )]##

    And if we assume :##\emptyset\neq \emptyset'##, then

    ##[\exists x(x\in \emptyset\wedge \neg x\in \emptyset')]\vee[\exists x(x\in \emptyset'\wedge \neg x\in \emptyset )]##.That implies :

    ##[ (x\in \emptyset\wedge \neg x\in \emptyset')]\vee[(x\in \emptyset'\wedge \neg x\in \emptyset )]##.Which in turn implies:

    ##[ (x\in \emptyset\vee x\in \emptyset')]##
     
  15. Nov 8, 2012 #14

    MLP

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    This argument is fallacious. You cannot infer that one and the same x is referred to by the two existential quantifers. If you could, it would be easy to prove that something is a Dodge and something is a Toyota implies that some one thing is both a Dodge and a Toyota. Maybe it would help to see this if you make the variable used in the second existential claim 'y'.
     
  16. Nov 9, 2012 #15

    Let : A= {1,2}, and B={1,2,3}.

    Can you prove that :## A\neq B## , using the axiom of extensionality??.

    Then you will find out that using the same or different quantifiers makes no difference
     
  17. Nov 9, 2012 #16

    MLP

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    I am not saying that the use of the different quantifiers makes a difference, it does not. I am saying that Existential Elimination does not allow you to assume that the x referred to in the first existential claim is one and the same x as the x referred to in the second existential claim.

    It is not legitimate to instanciate the first quantifier to x and then instanciate the second one to x and act like they are one and the same thing.
     
  18. Nov 9, 2012 #17

    Erland

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    Your argument shows that I was right, but perhaps that is what you meant?

    To be clear, you should have written the two last lines as

    ##\exists x[ (x\in \emptyset\wedge \neg x\in \emptyset')]\vee[(x\in \emptyset'\wedge \neg x\in \emptyset )]##

    and

    ##\exists x[ (x\in \emptyset\vee x\in \emptyset')]##.

    Then, the argument is correct, since

    ##\exists x(P(x)\vee Q(x))## and ##\exists x \,P(x)\vee\exists x\,Q(x)## are logically equivalent.
     
  19. Nov 9, 2012 #18
    Can you support that ,by writing a complete formal proof??

    Because i know that it will be useless to ask you ,where did you get the:

    "and ##\exists x[ (x\in \emptyset\vee x\in \emptyset')]##" ,part, e.t.c ,e.t.c
     
    Last edited: Nov 9, 2012
  20. Nov 10, 2012 #19

    Erland

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    Oops, the parenteses here are partially wrong. It should be:

    ##\exists x[ (x\in \emptyset\wedge \neg x\in \emptyset')\vee(x\in \emptyset'\wedge \neg x\in \emptyset )]##.
     
  21. Nov 10, 2012 #20

    Erland

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    Well, how did you get the corresponding line, in your previous post?

    If you want a complete formal proof, you must specify which formal system that should be used: which are the axioms and the rules of inference? Is a Hilbert style axiom system (and which variant in this case) or a natural deduction system (and which variant in this case) or some other kind of system?

    And whatever system is used, complete formal proofs tend to be extremely lengthy. One almost always takes shortcuts. But you have a habit of questioning all possible shortcuts.
     
    Last edited: Nov 10, 2012
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