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Prove a map of a space onto itself is bijective

  1. Sep 24, 2009 #1
    Hi,
    Say F:A->A where A is a metric space and F is onto. I think it should be true that this implies that F is also one to one. Is there a way to formally prove this? Thanks.
     
  2. jcsd
  3. Sep 24, 2009 #2

    Office_Shredder

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    Of course not. Let A be the set of natural numbers. Then F:A -> A by F(1)=1, F(2)=1, F(3)=2, F(4)=3,..F(k)=k-1

    And of course A is a metric space with the standard absolute value

    You need some other condition on F
     
  4. Sep 25, 2009 #3
    Oh I guess you're right. Ok if A and B are two metric spaces and there exists two onto functions F and G such that F:A->B and G:B->A, is there a way to prove that there exists a bijection mapping A to B? The reason I'm asking is because I'm trying to prove a comment from Rudin that says there is a bijection from the set of all Linear operators from R^n to R^m and the set of all real mxn matrices.
     
  5. Sep 25, 2009 #4

    Office_Shredder

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    The comment in Rudin doesn't really have anything to do with what you're talking about as far as I can tell.

    Given a linear map, it is determined entirely by how it maps the basis elements of Rn. And you can find a matrix that maps each basis element to any point of your choosing in Rm. So given a linear map, you can find a matrix which is the same function (and obviously every matrix is a linear map) and hence the two sets are essentially equivalent
     
  6. Sep 25, 2009 #5
    So linearity is the key to the proof? Or does a topological vector space being finite dimensional also play a role when it comes to being able to uniquely determine a linear mapping by how it maps basis elements?

    In other words, is the following statement true:

    Given a linear mapping L from a topological vector space X onto a topological vector space Y, L is "determined entirely by how it maps the basis elements of X?"

    I am looking at Functional Analysis "Big Rudin" page 16 Theorems 1.21 and 1.22 which relate local compactness of a topological vector space to that space necessarily having finite dimension.
     
  7. Sep 25, 2009 #6

    Office_Shredder

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    The finite dimensionality is important only for allowing you to actually write a matrix

    Of course. If you know how it maps the basis elements, then let v be in X.

    [tex]v= \sum_{i=1}^{n} \alpha_i v_i[/tex] for some basis vectors vi and field elements [itex] \alpha_i[/itex]. But we know precisely how to calculate
    [tex] L(\sum_{i=1}^{n} \alpha_i v_i)[/tex]

    Note that X and Y being topological has nothing to do with it
     
  8. Sep 27, 2009 #7
    Does every vector space have a basis?
     
  9. Sep 27, 2009 #8

    HallsofIvy

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    Yes. The fact that every finite dimensional vector space has a basis is one of the basic theorems of Linear Algebra. The fact that every infinite dimensional vector space has a basis requires something like Zorn's Lemma.
     
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