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Math Help - help with proving relaationship between sets

  1. #1
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    Exclamation help with proving relaationship between sets

    How would I prove this

    A⊆B ^ |A|=|B| -> A=B

    Is this proof ok and formal?

    Suppose (AB ^ |A|=|B|)
    Then |A|≠|B|
    Therefore A
    B ^ |A|=|B| -> A=B
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  2. #2
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    Quote Originally Posted by Sneaky View Post
    How would I prove this
    A⊆B ^ |A|=|B| -> A=B
    THIS IS ONLY TRUE OF FINITE SETS.

    That attempt falls short.
    Doing this by contradiction, suppose A\ne B~.
    But because A\subseteq B what does that imply?
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  3. #3
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    I forgot to mention that the sets were finite.
    So if A≠B then since A⊆B, then |A|<|B|
    therefore A⊆B ^ |A|=|B| -> A=B
    is this better?

    why wont it work for non-finite sets?
    Is it because then you cant say
    A⊆B ^ |A|=|B| ?
    Last edited by Sneaky; May 13th 2011 at 11:18 AM.
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  4. #4
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    Quote Originally Posted by Sneaky View Post
    why wont it work for non-finite sets?
    Is it because then you cant say A⊆B ^ |A|=|B| ?
    Let \mathbb{E} be the set even positive integers.
    Then \mathbb{E}\subsetneq \mathbb{Z}^+ BUT |\mathbb{E}|=|\mathbb{Z}^+|.
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  5. #5
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    does the symbol of the C with the 'not equal sign' under mean not a proper subset of ?

    And also is my proof ok now?
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  6. #6
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    does the symbol of the C with the 'not equal sign' under mean not a proper subset of ?
    No, it means that E is a proper subset of Z^+, i.e., it does not equal the whole Z^+.

    And also is my proof ok now?
    There is still a question whether enough details are provided to justify this statement: "if A≠B then since A⊆B, then |A|<|B|." This statement is true and sufficiently obvious, but whether more needs to be said depends on the course requirements and the definition of |A|.
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  7. #7
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    Quote Originally Posted by Sneaky View Post
    How would I prove this

    A⊆B ^ |A|=|B| -> A=B

    Is this proof ok and formal?

    Suppose (AB ^ |A|=|B|)
    Then |A|≠|B|
    Therefore AB ^ |A|=|B| -> A=B
    (1) As has been pointed out, we need to assume that either A or B is finite.

    (2) Your logic is wrong anyway. You don't prove P -> Q by assuming ~P and showing ~Q. You have it backwards. What you want here is to assume ~Q and show ~P.

    So suppose ~A=B. Show ~(A subset of B & |A|=|B|). Suppose A subset of B. Then, since ~A=B, we have A proper subset of B. But no finite set has the same cardinality as a proper subset of itself (this is called 'the pigeonhole principle').

    However, if you haven't already proven the pigeonhole principle, then you have to prove it, and that is a longer, more involved proof.
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  8. #8
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    i dont understand what more i can add to this:

    A and B are finite sets.
    if A≠B then suppose A⊆B, then |A|<|B|
    therefore A⊆B ^ |A|=|B| -> A=B
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  9. #9
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    Quote Originally Posted by Sneaky View Post
    if A≠B then suppose A⊆B, then |A|<|B|
    Prove it.
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