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Math Help - Set Theory

  1. #1
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    Set Theory

    Suppose S is finite of cardinality n. If f: S ---> T is a bijection, then T is finite and has cardinality n.

    Any advice on how to give a solid proof of this? THanks.
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  2. #2
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    Quote Originally Posted by jzellt View Post
    Suppose S is finite of cardinality n. If f: S ---> T is a bijection, then T is finite and has cardinality n.
    HINT: The composition of two bijections is a bijection.
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  3. #3
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    There's nothing to prove here, two sets have the same cardinality iff there's a bijection between the two sets.
    This is the definition.
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  4. #4
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    Quote Originally Posted by jzellt View Post
    Suppose S is finite of cardinality n. If f: S ---> T is a bijection, then T is finite and has cardinality n.

    Any advice on how to give a solid proof of this? THanks.

    From the book: General Topology,from Schaum's Outline Series i quote from page 32 the part that solves your problem:

    " A set A is called equivalent to a set B,written A~B,if there exists a function f: A --->B which is one to one and onto.The function f is then said to define a one to one correspondence between the set A and B.

    A set is finite iff it is empty or equivalent to {1,2,.........,n} for some nεN,otherwise it is said to be infinite.

    Clearly two finite sets are equivalent iff they contain the same number of elements.


    According to the above quotation ,since S is equivalent to T ,THEY have the same number of elements,and they have the same cardinality n
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