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Trouble w/ injections, surjections, bijections

  1. Oct 3, 2004 #1
    ok, from the definition, and drawing a picture, I can understand what all these mean. And when it comes to solving problems, I can solve them too. i.e 2x/1-x = 2y/y-1. easy, I know how to do it.

    But when it comes ot applying it to finite/infinite sets, I dont know how to start. i.e, is there a bijection from the positive real number to the set of natural numbers.

    ok, so positive real numbers imply 0, 1, 2,....n, and also everything in between.

    natural number imply 1, 2, ......n.

    so what does this mean? what does it mean if there is a bijection between these two? injection? surjection?

    how do I tink this through?
     
  2. jcsd
  3. Oct 4, 2004 #2
    Let P be the set of Positive real numbers and N is ofcourse the set of natural numbers

    Is there a function f:P->N such that every element of P is mapped to exactly one element of N and every element of N is an image of an element of P??

    Read through the definitions of function and injections , surjection and bijection and read through my statement ......

    Then once u have understood what i have said, give us your initial thoughts that is what do u think should be the answer to the question and why?

    -- AI
     
  4. Oct 4, 2004 #3

    matt grime

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    To be honest I don't think that's going to help much for this particular question. You're going to have to use some ingenuity to show that no map from N to R is a bijection.

    Any such map means you can list the real numbers x(1), x(2), x(3),... and get all of them in the list. Now you've got to show that there can be no such list by finding one not on the list. Hint, can you find a number different from x(n) in some way for all n?
     
  5. Oct 4, 2004 #4
    i wanted to skip the diagonal part atleast for the initial stages .... seeking an idea as to how much the person has thought upon the problem ...

    -- AI
     
  6. Oct 4, 2004 #5

    matt grime

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    1 isn't that obvious until you prove [0,1] is in bijection with R

    and

    3. is false.
     
  7. Oct 4, 2004 #6

    arildno

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    All right, message deleted.
     
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