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Need help to prove N is not infinite by induction

  1. Oct 20, 2009 #1
    1. The problem statement, all variables and given/known data
    Use induction to prove N is not finite.


    2. Relevant equations
    I think we just need to prove for all K in N, there is no bijection f:N-{1, 2, .....k}
    However, couldn't figure out. If there is anyone can help with, thanks a loooooot.

    3. The attempt at a solution
     
    Last edited: Oct 21, 2009
  2. jcsd
  3. Oct 20, 2009 #2

    Mark44

    Staff: Mentor

    What does N represent here? I'm guessing that it represents the natural numbers, of which there are an infinite number.
     
  4. Oct 20, 2009 #3
    That's right. N is natural numbers here. It is an infinite number. We need to prove that it is. Weird question, right? :(
     
  5. Oct 20, 2009 #4
    I'm guessing you misunderstood the question. Do you have the question as it is exactly stated in front of you?
     
  6. Oct 21, 2009 #5
    Re: Need help to prove N is not finite by induction

    I am sorry, I made a typo here. I mean N is not finite, use induction to prove that. Sorry.
     
  7. Oct 21, 2009 #6

    Office_Shredder

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    Can you prove there is no bijection from N to {1}?
     
  8. Oct 21, 2009 #7
    I sure can.

    Since f(1)=f(2)=f(3)=......=1, then f:N-{1} is not a one-to-one.
    So it is not a bijection.
     
  9. Oct 21, 2009 #8

    Office_Shredder

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    The inductive hypothesis will result in a similar argument. What have you tried so far?
     
  10. Oct 21, 2009 #9

    HallsofIvy

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    What is your definition of "finite"?
     
  11. Oct 21, 2009 #10
    When n=1, {1} it is easy to show that there is no bijection f:N-{1}

    suppose when n=K, there is no bijection f: N-{1, 2, .....k}, then I think I need to show that here is also no bijection f:N-{1, 2, 3, .....K , K+1}

    However, I couldn't figure out how to do that. I think I need to use contradtion to prove that. I tryed, however failed.

    : (
     
  12. Oct 21, 2009 #11
    I think this might be something:

    let S be a set

    let |N, the set of all natural numbers, be S U |N\S

    U = union symbol

    |N\S = the set of natural numbers excluding S

    for any n that is a natural number, n+1 is also a natural number, then n+1 is also an element of the natural numbers.

    for n = k, n+1= k+1
    If S: {1,......,k}, then |N\S is non empty, since there exists a K+1 which is also a natural number.
    so by the well ordering principle, |N\S has a least element, which is K+1.
    you can continue with this pattern for n = K+1 and n+1 = K+2 indefinitely to show that |N\S is never an empty set, so there are no finite sets S that can span all of |N
     
    Last edited: Oct 21, 2009
  13. Oct 21, 2009 #12
    I'm not sure if this will work. are you trying to show that for K there is no bijection implies K+1 has no bijection? if you prove this, then it will be true for all n that is a natural number.. which continues indefinitely into all natural numbers.. I don't know if you can get anywhere with this

    I think your focus should be using the idea that any set is finite and the natural numbers will always have more than any finite set - this makes it a non-finite set. I think the proof I posted above should work.. it's rough but I think that is the idea
     
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