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Suppose that f : R -> R is continuous and one-to-one

  1. Sep 14, 2006 #1

    ek

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    Sorry for no description of the type of question, but I'm so out to lunch on this stuff I don't even know what I'd call it. Ok, so I'm not looking for answers here, I'd just appreciate being pushed in the right direction.

    Suppose that f : R -> R is continuous and one-to-one (that is, x1 != x2 --> f(x1) != f(x2). Prove that for each interval I = [a,b] contained in R either f(I) = [f(a), f(b)] or f(I) = [f(b), f(a)].

    I have no idea what to consider for this problem.

    Sorry for the bad notation. I'm not too proficient at latex.
     
  2. jcsd
  3. Sep 14, 2006 #2
    Perhaps a proof by contradiction would work?
     
  4. Sep 14, 2006 #3

    HallsofIvy

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    Staff Emeritus
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    It is hard to suggest how to prove something like that without knowing what you have to work with. For example, there is a theorem saying that a continous function maps an interval into an interval. Do you know that theorem and can you use it? It's a special case of a more general theorem that says that a continous function maps a connected set into a connected set- and, of course, in R connected sets are precisely intervals.

    If you can use that theorem then you have immediately that f([a, b])= [c, d] for some c, d. Then use "one-to-one" to prove c= f(a), d= f(b) or c= f(b), d= f(a).
     
  5. Sep 15, 2006 #4

    ek

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    Thanks very much for the help and sorry for the late reply.

    This is a third year "advanced calculus" class and I was privy to the connected set theorem.

    I suck (relatively anyway, compared to phys/astr) at math, getting through this final major math class will be quite a challenge to get my degree (Astronomy).

    I'll probably be calling upon the very knowlegable PF members for some guidance throughout the semester.

    Thanks again.
     
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