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Continuous functions

  1. Dec 13, 2009 #1
    1. The problem statement, all variables and given/known data
    Suppose [tex]f: [0,1] \rightarrow R [/tex] is two-to-one. That is, for each [tex] y \in R, f^{-1}({y}) [/tex] is empty or contains exactly two points. Prove that no such function can be continuous.


    2. Relevant equations
    Definition of a continuous function:
    Suppose [tex]E \subset R[/tex] and f: E \rightarrow R[/tex]. If [tex]x_0 \in E[/tex] then f is continuous at [tex]x_0[/tex] iff for each [tex]\epsilon > 0[/tex], there is [tex]\varphi >0[/tex] such that if
    [tex]|x - x_0| < \varphi, x \in E[/tex],​
    then
    [tex]|f(x) - f(x_0)| < \epsilon[/tex].​

    If f is continuous at x for every [tex]x \in E[/tex], then we say f is continuous.


    3. The attempt at a solution
    I think from an algebraic view, there is more elements in the domain then codomain which means that the function f is not one-to-one, but onto. Though I know this fact, I am pretty sure this will not aid in my attempt to prove this problem. Could someone help me understand why f cannot be continuous.


    Thanks,


    Jeffrey Levesque
     
    Last edited: Dec 13, 2009
  2. jcsd
  3. Dec 14, 2009 #2
    Try assuming that f is continuous and then using the extreme and intermediate value theorems to produce a y in R with more than two values in its pre-image.
     
  4. Dec 14, 2009 #3

    HallsofIvy

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    Staff Emeritus
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    Prove, rather, that there exist y such that [math]f^{-1}(\left{y\right})[/math] contains exactly one value.
     
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