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Prove ΦxΦ =Φ

  1. Sep 3, 2009 #1
    Prove:

    [tex] \emptyset\times\emptyset =\emptyset[/tex]
     
  2. jcsd
  3. Sep 4, 2009 #2
    The empty set has no elements. Therefore, there is no ordered pair of elements of the empty set (using any of the common definitions of an ordered pair). There isn't really much to prove.
     
  4. Sep 4, 2009 #3

    HallsofIvy

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    Or, if you want to get "fancy", a proof by contradiction: Suppose that [itex]\Phi \times \Phi[/itex] is not empty. Then there exist a pair, (a, b) in [itex]\Phi \times \Phi[/itex] such that [itex]a\in \Phi[/itex] which contradicts the fact that [itex]\Phi[/itex] is empty.
     
  5. Sep 4, 2009 #4
    Wrong, no theorem to base your contradiction.
     
  6. Sep 4, 2009 #5

    jgens

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    I'm not a mathematician, so perhaps my comment is naive, but the empty set is defined as the unique set which contains no elements. If [itex]a[/itex] were a member of the empty set, then the empty set would not be empty which is a contradiction. Maybe I'm misunderstanding something here?
     
  7. Sep 4, 2009 #6
    You're not. Evagelos is wrong in assuming that there is no contradiction. It contradicts (assuming we're talking about ZF) the axiom of the empty set.
     
  8. Sep 4, 2009 #7

    Hurkyl

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    :confused: HoI's statement is a theorem.
     
  9. Sep 4, 2009 #8

    statdad

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    Halls' approach is, as others have stated, fine.
     
  10. Sep 4, 2009 #9
    Yeah, I thought that was pretty much the definition of cartesian product: If A and B are sets, and C = AxB, then y is in C iff there is an a in A and a b in B such that y = (a, b).

    If one of A or B is the empty set, then for every y, it is not in C.
     
  11. Sep 4, 2009 #10
    Let Hall write a formal proof then to find out whether his proof is right or wrong.

    Sorry Halls i know that you dislike formal proofs,but that is the only way,otherwise we will argue indefinitely
     
  12. Sep 4, 2009 #11
    No ,i did not say that there is no proof by contradiction,but Hall's contradiction is not based somewhere
     
  13. Sep 4, 2009 #12
    Like I said, his contradiction is based on the axiom of of the empty set, is it not?

    What do you mean by "based on" then?
     
  14. Sep 5, 2009 #13

    CRGreathouse

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    The proof is so shockingly obvious that I'm surprised even you could ask for it. I can only assume this is a homework problem.
     
  15. Sep 7, 2009 #14

    CompuChip

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    I fail to see how Hall's proof is not a formal one.

    Alternatively, you could give a direct proof: it is true that
    [tex] \emptyset\times\emptyset \subseteq \emptyset [/tex]
    because for all [itex]x \in \emptyset\times\emptyset[/itex], [itex]x \in \emptyset[/itex] - similarly the other inclusion follows. (In fact, if [itex]A \subseteq \emptyset[/itex], then it is always the case that [itex]A = \emptyset[/itex]).
     
  16. Sep 7, 2009 #15

    HallsofIvy

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    We have asked evagelos repeatedly what he means by a "formal" proof and he has never answered clearly. I suspect he means something like used in "Principia Mathematica".
     
  17. Sep 7, 2009 #16

    CompuChip

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    Wait, let me get my ruler and compass...
     
  18. Sep 7, 2009 #17
    I suppose if you wanna get fancy, you can always try a categorical proof (of isomorphism, that is).
     
    Last edited: Sep 7, 2009
  19. Sep 9, 2009 #18
    Here is a formal proof of [tex] \emptyset\times\emptyset = \emptyset[/tex],a direct one:


    1)[tex]\forall A\forall B [ A\times B =\emptyset\Longleftrightarrow A=\emptyset\vee B=\emptyset][/tex].....................................................................................a theorem in set theory


    2) [tex]\emptyset\times\emptyset =\emptyset\Longleftrightarrow \emptyset = \emptyset\vee\emptyset =\emptyset[/tex]..............................................................................from (1) and using Universal Elimination ,where we put [tex] A=\emptyset[/tex] and [tex] B=\emptyset[/tex]

    3)[tex]\emptyset=\emptyset\vee \emptyset =\emptyset\Longrightarrow \emptyset\times\emptyset =\emptyset[/tex]...............................................................................from (2) and using biconditional Elimination


    4)[tex]\forall x[ x=x][/tex]...........................................................................................................an axiom in equality


    5)[tex]\emptyset =\emptyset[/tex].............................................................................................................from (4) and using Universal Elimination,where we put [tex] x=\emptyset[/tex]


    6)[tex]\emptyset =\emptyset\vee \emptyset =\emptyset[/tex]......................................................................................from (5) and using idempotent law ( p ====> p or p)


    7)[tex]\emptyset\times\emptyset =\emptyset[/tex]....................................................................................from (3) and (6) and using M.Ponens.

    Now if you wish, write a formal proof using contradiction
     
  20. Sep 10, 2009 #19

    statdad

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    Okay (haven't looked through this, right now will take you on face value) but that does not disqualify Hall's proof as being perfectly valid.
     
  21. Sep 10, 2009 #20

    CRGreathouse

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    Yeah, with theorems like this it's pretty easy.
     
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