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Equal sets and bijective correspondence

  1. Dec 7, 2011 #1
    1. The problem statement, all variables and given/known data

    If [n] and [m] are equal, then they are bijective correspondent.

    I define [itex]f \subset\{(n,m)\mid n \in [n], m\in [m]\}[/itex]. Suppose [n]=[m]. Let[itex](n,m_1),(n,m_2)\in f.[/itex] Because [n]=[m], then [itex]m_1=m_2[/itex]. So for all [itex]n \in [n][/itex], there exists a unique [itex]m\in [m][/itex] such that f(n)=m. So f is a function.

    Next I want to prove f is surjective and injective. But I'm stuck. How can I make use of the supposition [n]=[m] to prove surjectivity and injectivity?

    Thanks.
     
  2. jcsd
  3. Dec 8, 2011 #2

    HallsofIvy

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    Do you mean "equal" or "equivalent"? In order to be equal, sets A and B must be identical- they contain exactly the same elements. In order to be equivalent, they must have the same cardinality- contain exactly the same number of elements. In either case, there is an obvious bijection.
     
  4. Dec 8, 2011 #3
    I mean equal - so both sets contain the same elements.

    Yeah, there is obviously a bijection. But how can I show that using a rigorous mathematical proof?
     
  5. Dec 8, 2011 #4

    HallsofIvy

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    The obvious way- show that f, from [m] to [n], defined by f(x)= x, is both "one-to-one" and "onto". And that should be easy- if f(a)= f(b), what must be true of a and b? Let c be a member of [n]. What member of [m] is mapped to c?
     
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