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Continuity of a function

  1. Nov 20, 2008 #1
    1. The problem statement, all variables and given/known data
    Suppose f: [0,1] -> [0,1] is such that f attains each of its values exactly twice
    Show that f cannot be continuous


    3. The attempt at a solution
    I assumed that f is continuous and tried to break it up into cases and show that there must be a value that is obtained 3 times.
    since f is defined on an interval it has a sup and an inf (each attained twice by hypothesis). my cases are the order in which the are attained, i.e. 1) max,min,max,min or 2) max,max,min,min
    case 1 is easy, but i can't figure out how to do case 2
    is this the right approach or is there an easier way?
     
    Last edited by a moderator: Nov 20, 2008
  2. jcsd
  3. Nov 20, 2008 #2

    Vid

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    Are you sure you phrased the question right? Your sentence ends abruptly and the hypothesis you used in the proof doesn't match the hypothesis you gave.
     
  4. Nov 20, 2008 #3
    my bad. poor mistake for me to make
     
  5. Nov 20, 2008 #4

    Dick

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    kfrob, stick to your guns and don't just admit you are wrong when you aren't. If you have either (min,max,min,max) or (max,min,max,min) if min<y<max how many times is f(x)=y if f is continuous? At least. Use the IVT. If it's (max,max) what does that mean?
     
  6. Nov 20, 2008 #5
    i got that part, but i can't figure out how to show that f attains a given value 3 time if you have (max,max,min,min) or (min,min,max,max). drawing it i can see that is true, but any heuristic argument involves me saying that f has a "relative min" between the maximums and a "relative max" between the minimums. IVT is about as far as we've gotten so i can't make any argument about f'. is there a way to get around this?
     
  7. Nov 20, 2008 #6

    Dick

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    If you have two different consecutive maxes, doesn't there have to be a min in between? Otherwise the function must be constant in between? Which means it achieves that value an infinite number of times?
     
  8. Nov 20, 2008 #7
    i finally got it. thank you
     
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