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For sets A,B, if I can that show |A|=|B| and A is a subset of B, then A=B?

  1. Jun 25, 2012 #1
    [Note: If this is posted in the wrong forum, I'm very sorry. It is directly related to a textbook question.]

    This may be a silly question.

    I know that I can prove two sets to be equal by showing that they are subsets of each other. But, what if I have that two sets have the same cardinality and that one set is a subset of the other. Is this enough to show that the sets are indeed equal?

    Thanks.
     
  2. jcsd
  3. Jun 25, 2012 #2

    micromass

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    Think about [itex]A=\mathbb{N}[/itex] and [itex]B=\mathbb{Z}[/itex].
     
  4. Jun 25, 2012 #3
    You're right!! :)

    But, I forgot to mention that they are finite. :(
     
  5. Jun 25, 2012 #4

    micromass

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    OK, then it is indeed true. Do you need to prove this?? It's pretty obvious, but the proof might be quite complicated.
     
  6. Jun 25, 2012 #5
    Suppose a ε <b>. Then <a> = <b> iff a and b have the same order.

    Proof [just the converse] :
    Let G be a group. Let a,b be elements of G. Suppose a ε <b> and suppose a and b have the same order. Let n=ord(a)=ord(b). Then, the cyclic groups generated by a and b have the same order. That is, n=|<a>|=|<b>|. Now, from our assumption, a ε <b> and, then, since <b> consists of all powers of b, a = bk for some integer k. Hence, <a>=<bk>, which is a subset of <b>. Then since <a> is a subset of <b> and they have the same cardinality, they must be equal. Hence, <a>=<b>, our desired result. QED
     
  7. Jun 25, 2012 #6

    micromass

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    Yeah, that seems like a perfectly valid proof!!
     
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