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One to one functions

  1. Jul 5, 2005 #1
    1. use the inverse function theorem to prove that any function [tex] f: R^2 -> R [/tex] cannot be one-to-one. hint: let
    g(x,y) = (f(x,y), y) near an appropriate point.

    2. prove #1 using the implicit function theorem.

    3. generalize part 2) to show that no function
    [tex] f:R^n -> R^m, [/tex] with n>m can be one-to-one.


    i am not sure where to start on this problem. for #1, i don't know how to apply the hint that's given. what is this "appropriate point" it's referring to? i guess i started by trying to take the determinant of f'(x,y). but f'(x,y) is a 2x1 matrix...how do i take determinant of that? not sure where this is leading me.


    not sure what to do for #2 or #3 either. my understanding of the implicit function theorem is very fuzzy. like, i understand how to use it to differentiate one variable with respect to another, but i don't know how to use it to prove the function is not one to one. any help is appreciated - thanks in advance.
     
  2. jcsd
  3. Jul 5, 2005 #2

    AKG

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    I guess you mean something like "any continuous function..."?
     
  4. Jul 5, 2005 #3
    yes, f is continuous
     
  5. Jul 6, 2005 #4

    matt grime

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    start by writing out the statement of the inverse function theorem. and remember if differentiable a function fro R to R is not 1-1 then there exist poitns x and y with g(x)=g(y), so what can we say about the derivative somewhere in the middle?
     
  6. Jul 7, 2005 #5

    saltydog

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    You know what, I can't let this one slip into oblivion down there without making a final comment:

    I've been working with this one and have come to the conclusion that it represents a fundamental feature of euclidean geometry, and hense of our world and therefore places boundaries on the types of transformations we can observe in nature.

    I am thus immediately led, perhaps foolishly, to the suspicion that other geometries such as non-euclidean ones may not share this feature, that is, a one-to-one mapping under equivalent conditions would be possible.

    I'll keep working with it. :smile:
     
  7. Jul 7, 2005 #6

    matt grime

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    this is nothing to do with geometry and is entriely down to topology. with the diiscrete topology R^2 and R are homeomorphic.

    (what other geometries on R^2 are you thinkig of anyway?
     
  8. Jul 7, 2005 #7

    saltydog

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    I suppose I find it interesting that no matter how I attempt to bend or twist or deform a surface above the x-y plane, I will not be able to make it 1-to-1 such that if (x,y)[itex]\ne[/itex] (r,s) then f(x,y)[itex]\ne[/itex]f(r,s). Is it just the nature of eclidean geometry in 3-D space that prevents such? Can a geometry be constructed that allows 1-to-1 when n>m?

    I realize one implication of the Implicit Funtion Theorem is that in this case, the number of independent variables, (n), is more than the number of equations, (m). Thus an infinite number of solutions is obtained and only when n=m can we obtain uniqueness and thus a unique inverse (at least that's how I'm interpreting it).

    Still, I just can't help imagining something intrinsic about the geometry of space somehow constraining this. Interesting . . .
     
  9. Jul 7, 2005 #8

    matt grime

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    it's the metric, it's the topology that make this true. agreed the metric is that inherited from the euclidean norm, but it needn't be since any two normed topologies are equivalent on a finite dimensional vector space. wwith a differenttoplogy there is a continuous 1-1 map from R^2 to R eg any bijection in the discrete topology.
     
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