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Prove that if S is a subset of A then S is an empty set.

  1. May 11, 2012 #1
    1. The problem statement, all variables and given/known data
    Let S be a set such that for each set A, we have S[itex]\subseteq[/itex]A. Show that S is an empty set.

    3. Relevant equations
    10.6 Proposition. For each set A, we have empty set [itex]\subseteq[/itex] A.

    3. The attempt at a solution
    Solution. Consider any set A and a set S such that S[itex]\subseteq[/itex]A. Choose any x[itex]\in[/itex]S. Then since S[itex]\subseteq[/itex]A we also have x[itex]\in[/itex]A. From proposition 10.6 we know that if x[itex]\subseteq[/itex]empty set, then x[itex]\subseteq[/itex]A. Now x is arbitrary. Thus S must be an empty set.

    I feel like this proof is horrible and doesn't flow at all. Can someone give me some pointers?
     
  2. jcsd
  3. May 11, 2012 #2

    jgens

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    Clearly [itex]\emptyset \subseteq S[/itex] and by hypothesis [itex]S \subseteq \emptyset[/itex]. Can you put this together?
     
  4. May 11, 2012 #3
    I am trying to see how this works but I don't get it.

    I understand that if we can show that [itex]\emptyset \subseteq S[/itex] and [itex]S \subseteq \emptyset[/itex], then we can say that [itex]S=\emptyset[/itex] but I don't understand what you mean by, by hypothesis [itex]S \subseteq \emptyset[/itex].
     
  5. May 11, 2012 #4
    Nevermind. Beat to the punch.
     
  6. May 11, 2012 #5

    jgens

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    In your own words: Let S be a set such that for each set A, we have S⊆A. The empty set is a set, yes?
     
  7. May 11, 2012 #6
    Let me apologize for being a bit slow with grasping this. Ever since my proof class moved into sets I've been a little slow...

    Anyhow, from the way I am thinking about it, the problem is to show that there exists a set such that it is a subset of A. Then by saying that [itex]S\subseteq\emptyset[/itex] we are claiming that [itex]\emptyset[/itex] is such a set. Is that the correct way of looking at it?
     
  8. May 11, 2012 #7

    jgens

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    I can't make sense out of what you're saying, so could you be a little more clear?
     
  9. May 11, 2012 #8
    I can't see how saying Let S be a set such that for each set A, we have S⊆A can lead to [itex]S\subseteq\emptyset[/itex].
     
  10. May 11, 2012 #9

    jgens

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    That is clearer, thank you. Do you accept that [itex]\emptyset[/itex] is a set?
     
  11. May 11, 2012 #10
    Yes I do.
     
  12. May 11, 2012 #11

    jgens

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    Since [itex]S[/itex] is a subset of every set and the empty set is a set, this mean [itex]S \subseteq \emptyset[/itex] (i.e. S is a subset of the empty set). Not sure what else to say about that. If you're still confused perhaps another member could chime in and help?
     
  13. May 11, 2012 #12
    Ok. Here is an attempt at a better proof.

    Consider any set A and a set S such that S is a subset of A. From proposition 10.6 we know that the empty set is a subset of S. Now S is a subset of every set A so it follows that S is a subset of the empty set. Thus we have the empty set is a subset of S and S is a subset of the empty set. Therefore S is an empty set.

    Does that work?
     
  14. May 11, 2012 #13

    jgens

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    Yeah. The only thing I would change is that I would say S is the empty set. The empty set is provably unique.
     
  15. May 11, 2012 #14
    Done.

    Thanks so much for your help! I need to stare at this till it makes more sense to me.
     
  16. May 11, 2012 #15

    jgens

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    Sometimes that is the only approach, haha. If you still have difficulty with the [itex]S \subseteq \emptyset[/itex] feel free to ask the other members. I just don't know how else to explain it :/
     
  17. May 11, 2012 #16
    There's only one empty set. That's because a set is completely characterized by its elements. There's only one set with no elements, that's the empty set.
     
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