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Proving that a function can't take exactly same value twice

  1. Apr 13, 2016 #1
    1. The problem statement, all variables and given/known data
    Prove that there does not exist a continuous function f, defined on R which takes on every value exactly twice.

    2. Relevant equations
    It uses this property:
    1... If f is continuous on [a,b], then there exists some y in [a,b], such that f(y)≥f(x), for all x in [a,b]


    3. The attempt at a solution
    This problem was taken from Spivak-calculus.
    He did give a hint in the problem: He asks us to consider f(a) for some 'a' in R. By definition, the function takes the same value at some b in R. So, f(b)=f(a).
    I am considering the case when x is in [a,b] and f(x)≥f(a). For all other x excluding this interval, f(x)<f(a)
    f is continuous on R, which means that it must be continuous on [a,b]. By using the above property,,there exists some y in [a,b], such that f(y)≥f(x), for all x in [a,b]. But by definition of f, there is a y1 in [a,b] such that f(y)=f(y1). Here y1<y or y1>y.....let us denote set D=[y,y1] or [y1,y] depending on which is greater(either y or y1). Here, let D=[y,y1]
    Let us consider D. f, is continuous on D. so, there exists an k in D such that f(k)≤f(m) for all m in D.
    There exists no h in D such that f(h)=f(y). Orelse, the function takes three values....a clear contradiction.
    But k is in [a,b], so this means that f(k)≥f(a). But clearly f(k) is not equal to f(a).
    So, f(k)>f(a) but less than f(y).
    since f(y)>f(k)>f(a) and f is continuous on [a,y], there exists some p in [a,y] such that f(p)=f(k)'
    Also in [y1,b], we have f(y1)>f(k)>f(b) and f is continuous. so there exists z in [y1,b] such that f(z)=f(k1)
    We have three points(z,p,k) where the value of the function is same, a contradiction to the assumption of our f. Hence, there exists no f which takes the same value twice in R. Hence proof. Is my proof right?
     
  2. jcsd
  3. Apr 13, 2016 #2

    stevendaryl

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    That doesn't seem right. You have that [itex]f(a) = f(b)[/itex]. So somewhere between [itex]a[/itex] and [itex]b[/itex], the function has its maximum value for that interval. So you're assuming that that is at the point [itex]y[/itex]. So [itex]a \leq x \leq b \Rightarrow f(x) \leq f(y)[/itex]. Now, you say:

    "...by definition of [itex]f[/itex], there is a [itex]y_1[/itex] in [itex][a,b][/itex] such that [itex]f(y) = f(y_1)[/itex]"

    That's not right. All you know is that there is some [itex]y_1[/itex] such that [itex]f(y) = f(y_1)[/itex]. You don't know that [itex]y_1[/itex] is in the interval [itex][a,b][/itex]
     
  4. Apr 13, 2016 #3
    We have taken f such that
    f <f(a), for all x<a and x >b.
    So, y1 should lie this interval, if my assumption is not wrong.
     
  5. Apr 13, 2016 #4

    stevendaryl

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    Oh. I thought the only thing you were assuming was that [itex]f(a) = f(b)[/itex]. How do you know you can assume that [itex]f(x) < f(a)[/itex] for [itex]x < a[/itex]?
     
  6. Apr 13, 2016 #5
    Let f(x)≥f(a) for x<a and x>b,
    Consider the case when x<a. There is a point j present such that j<a,
    The function f is contiunous at [j,a].
    Here f(x)≥f(a), for all x in [j,a]
    Thus the minimum value of f in this interval is f(a).
    By definition of f, again there is f(b)=f(a) for some b.
    for the interval [a,b], again f has some x in [a.b] such that f(x)≥f(a).
    Considering the case x<b, we can easily prove that f takes 4 values for some points x in R. Hence a contradiction.
     
  7. Apr 13, 2016 #6

    stevendaryl

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    But why can you assume that?

    I'm not saying you're wrong, only that you're missing a step in the reasoning.

    Here's the step that you're missing (or it seems that way to me): If you look at a graph of [itex]f(x)[/itex] versus [itex]x[/itex], and draw a horizontal line through the point [itex]x=a, y=f(a)[/itex], you know that the graph can only cross that line twice, at [itex]x=a[/itex] and [itex]x=b[/itex]. That means that the graph never crosses the line in the region [itex]x < a[/itex]. So that means either
    1. [itex]f(x) > f(a)[/itex] throughout the whole region, or
    2. [itex]f(x) < f(a)[/itex] throughout the whole region.
    (You can reason similarly for the region [itex]x > b[/itex]).

    So you can do a case split between the two possibilities for [itex]x < a[/itex].
     
  8. Apr 14, 2016 #7
    Yeah, this broader intuition is more comprehendable.
     
  9. Apr 15, 2016 #8
    Is the proof right?
     
  10. Apr 15, 2016 #9

    stevendaryl

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    I think that under your assumption that [itex]f(x) < f(a)[/itex] in the regions [itex]x < a[/itex] and [itex]x > b[/itex], your proof is correct, but you need to prove it without making that assumption.
     
  11. Apr 15, 2016 #10
    So, is there any way of proving it?
     
  12. Apr 15, 2016 #11

    stevendaryl

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    You sort of have the right idea. But first you have to consider the possibilities outside the range [itex][a,b][/itex].

    You can show that
    1. either [itex]f(x) < f(a)[/itex], for all [itex]x < a[/itex], or [itex]f(x) > f(a)[/itex], for all [itex]x < a[/itex]
    2. either [itex]f(x) < f(a)[/itex], for all [itex]x > b[/itex], or [itex]f(x) > f(a)[/itex], for all [itex]x > b[/itex].
    So there are 4 cases:
    1. [itex]f(x) < f(a)[/itex] for all [itex]x < a[/itex], and [itex]f(x) < f(a)[/itex] for all [itex]x > b[/itex]
    2. [itex]f(x) < f(a)[/itex] for all [itex]x < a[/itex], and [itex]f(x) > f(a)[/itex] for all [itex]x > b[/itex]
    3. [itex]f(x) > f(a)[/itex] for all [itex]x < a[/itex], and [itex]f(x) < f(a)[/itex] for all [itex]x > b[/itex]
    4. [itex]f(x) > f(a)[/itex] for all [itex]x < a[/itex], and [itex]f(x) > f(a)[/itex] for all [itex]x > b[/itex]
    So you have to show that each of these leads to a contradiction. You have a proof of case 1, but you have 3 other cases.
     
  13. Apr 15, 2016 #12
    Now I get it. You want me to consider all the possible cases. Thank you.
     
  14. Apr 15, 2016 #13

    PeroK

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    Choosing ##a, b## such that ##f(a) = f(b)## was a good start. I would then have drawn a graph and used the logic of what sort of graph you can and can't draw to guide your proof.
     
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