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Is my prove correct?

  1. Mar 13, 2008 #1
    There is a simple problem, and I gave my simple prove. Could anybody help me check whether it is correct:

    Show that if B is not finite and [tex]{B}\subset{A}[/tex], then A is not finite.

    My prove:
    Since B is not finite, there exists a bijection of B into one of its proper subset, C, say, and denote the function to be f: [tex]{B}\rightarrow{C}[/tex].
    Since B is a proper subset of A, as an extension of f to A,there exist a (some) bijiection(s), say g, such that g: [tex]{A}\rightarrow{C}[/tex]. And thus g maps A onto its proper subset, A cannot be finite. Hense A is infinite.

    Thks!
     
    Last edited: Mar 13, 2008
  2. jcsd
  3. Mar 13, 2008 #2

    HallsofIvy

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    What, precisely, do you mean by "an extension of f to A". In other words, what, exactly, is g?
     
  4. Mar 13, 2008 #3

    Dick

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    You should also notice that there doesn't generally exist a bijection of A to C. Which is ok, because to show A is infinite you don't need a bijection with C, you only need one with 'some subset' of A.
     
  5. Mar 13, 2008 #4
    can't you simply say if [tex]{B}\subset{A}[/tex]

    n [tex]\le[/tex] cardinality of (B) [tex]\leq[/tex] cardinality of (A) for all n in N.

    So A is not finite.
     
    Last edited: Mar 13, 2008
  6. Mar 13, 2008 #5

    Dick

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    Yes, if you have cardinality theorems available. Ka Yan appears to want to prove it using the definition that a set is infinite if there exists a bijection with a proper subset of itself.
     
  7. Mar 14, 2008 #6
    Yes, thanks gentlemen.

    That, I was originally considered, if f is a function on a subset B of A, then an "extension" of f from B to A is that, exist a function g on A, such that g(x)=f(x) for all x in B.

    But B and A are seem neccessary to be closed, and I didnt quite sure if those A and B are closed or not to asure the concept "extension" be apply.

    Besides, my classmate gave me another way:
    construct a sequense {an} by: take a0[tex]\in[/tex]B, a1 [tex]\in[/tex] B-{a0}, a2 [tex]\in[/tex] B-{a0, a1}, and so on. And this sequense is infinite, and all belong to A, thus A is infinite.
    I don't quite well remember the constructing process, and I showed just what he meant.
     
  8. Mar 14, 2008 #7

    Dick

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    Here's another way. Consider the subset D=(A-B) union C. It's a proper subset of A since C is a proper subset of B. Define g(x)=f(x) for x in B and g(x)=x for x in A-B. g is a bijection of A with a proper subset D. So A is infinite. I think that's the 'extension' you were after.
     
    Last edited: Mar 14, 2008
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