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Proving a function is disontinuous everywhere

  1. Oct 8, 2008 #1
    1. The problem statement, all variables and given/known data

    Let f(x) be continuous in the closed interval [0,1]. Suppose further that f(x) assumes rational values only and that f(x) = 1/2 when x = 1/2. Prove that f(x) = 1/2 everywhere.

    2. Relevant equations

    N/A

    3. The attempt at a solution


    If f is continuous at x = 1/2, then for every positive e (epsilon), we can find a positive number d (delta) such that |f(x) - f(1/2)| < e for all x in the domain of f such that |x-1/2| < d. Now, we have f(1/2) - e < f(x) < f(1/2) + e. F assumes rational values only, but we know that between any two rationals numbers, there are an infinite amount of irrational numbers, so there is an irrational value, say c in the open interval
    f(1/2) - d < x < f(1/2) + d. Since the function is continuous, we will take a smaller epsilon, say e1, with e1 < e. This is done as to "get rid" of the irrational. However, again by the density of the irrationals in the reals, there is another irrational value in that even smaller epsilon-neighborhood. In an interval length of 2en, where en > 0, we can always find an irrational in this interval. So we msut take e = 0 and the inequality becomes
    f(1/2) = 1/2 < f(x) < f(1/2) = 1/2, meaning that f(x) = 1/2. QED

    For starters, I'm not sure if this proof is correct. Also, would the inequality 1/2 < f(x) < 1/2
    need to be changed to 1/2 <= f(x) <= 1/2 in order for it to imply that f(x) = 1/2?
     
  2. jcsd
  3. Oct 8, 2008 #2

    Dick

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    No, the proof's not very convincing. Look at the Intermediate Value Theorem.
     
  4. Oct 8, 2008 #3
    What if I use a contradiction and argue in a similar way? Like this:

    Suppose that f(x) > 1/2 such that f(x) does not take on any irrational values for all x values not equal to 1/2. Then since the function is continuous , |f(x) - f(1/2)| < e, implying that 1/2 - e < f(x) < 1/2 + e. By the IVT, there must be an irrational value, c, in (1/2 - d, 1/2 + d) such that
    1/2 - e < f(c) < 1/2 + e. However, we said that f cannot take on irrational values. So f(x) cannot be greater than 1/2. Assume now that f(x) < 1/2. Again by the IVT there is an irrational value c in (1/2-d,1/2+d) such that 1/2 - e < f(c) < 1/2 + e. Again this is a contradiction. Since f(x) cannot be greater than or less than 1/2, then f(x) must be equal to 1/2. QED
     
  5. Oct 8, 2008 #4

    Dick

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    It's not nearly that complicated. If f(a)=c>1/2, then f(x) takes on all values in the range [1/2,c] for x in the interval between a and 1/2. No e's or d's needed.
     
  6. Oct 8, 2008 #5
    Is the proof still correct though?
     
  7. Oct 8, 2008 #6

    Dick

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    No. It doesn't quite make sense. Usually when talking about continuity you say thinks like "for all e>0 there exists a d>0 such that..." around the point of continuity. x might not be terribly close to 1/2. Put it this way, I don't understand it.
     
  8. Oct 8, 2008 #7
    Let me try to reword it:

    Suppose that f(c) > 1/2 for some c in [0,1]. Now, by the IVT, for every x in the interval [1/2,c], there is a corresponding y value in the interval [1/2,f(c)]. Since it takes on every value between [1/2,c], then it must take on an irrational value, say f(k), in this interval. However, f cannot take on any irrational values, so it cannot be greater than 1/2. By this same argument, it also cannot be less than half. So, in the interval [0,1], f(x) = 1/2.

    I can't think of a way to prove it is equal to 1/2 everywhere (unless the question is really only asking to prove it equals 1/2 in the domain [0,1]).
     
  9. Oct 8, 2008 #8

    Dick

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    That's more like it. Sure, they only want you to prove it on [0,1]. That's the domain of the function. But you could have proved it on any interval containing 1/2. There is one factual inaccuracy in the proof, "Now, by the IVT, for every x in the interval [1/2,c], there is a corresponding y value in the interval [1/2,f(c)]." You don't know for every x in [1/2,c] that f(x) is in [1/2,f(c)]. The function could go outside those bounds. But you do know for every y in [1/2,f(c)] that there is an x (that's the IVT). That's all you need.
     
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