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Two-to-One Function

  1. Jun 10, 2010 #1
    Hi, I've been trying to prove this statement for a while now but haven't made much progress:

    Suppose [tex]f:[a,b]\to\mathbb{R}[/tex] has the property that for each [tex]y[/tex] in the image of [tex]f[/tex], there are EXACTLY two distinct points [tex]x_1,x_2\in [a,b][/tex] that map to it. Then, [tex]f[/tex] is not continuous.

    Well, I considered two approaches. I figure it has to do with the intermediate value theorem somehow, but I could be wrong. The argument that I'm hoping would do the trick goes as follows and it is a proof by contradiction approach:

    We shall assume that [tex]f[/tex] is continuous. Construct a function [tex]g:[a,b]\to[a,b][/tex] as follows. Each [tex]x\in [a,b][/tex] has a unique partner [tex]x'\in[a,b][/tex] such that [tex]f(x)=f(x')[/tex] via the two-to-one property of [tex]f[/tex]. So, we can have [tex]g:x\mapsto x' [/tex] and [tex]g[/tex] is also bijective. Now, here's the problem. I have an instict that [tex]g[/tex] is continuous because of [tex]f[/tex] but I don't know how to prove it. If this is indeed true, then we're done because of the fixed point theorem.

    Any suggestions would be appreciated, thanks:)
    Last edited: Jun 11, 2010
  2. jcsd
  3. Jun 11, 2010 #2


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    Suggestion: [a,b] can be divided into two intervals L and U as follows. For each pair (x1,x2) that have the same image and x1 < x2, put x1 in L and x2 in U. I suspect all points in L are then < all points in U (you will need to use the fact that there are exactly two points going into the image). Since they must make up the entire interval [a,b], L and U will meet at some point x. There will be no other point x' such that f(x)=f(x').
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