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Math Help - Continuity

  1. #1
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    Continuity

    I need to show that if f is a function with domain on the closed interval [a,b] and the range of f is {-1,1}, then there is a number x in [a,b] at which f is not continuous.

    I'm guessing it is not continuous at a or b or both but I'm not sure how to go about proving it.
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  2. #2
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    If the range is {-1,1}, you can find x,y in [a,b] such that f(x)=-1 and f(y)=1. f can't be continuous in the interval [x,y] (or [y,x]!) because of Bolzano.
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  3. #3
    Super Member redsoxfan325's Avatar
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    I would have gone with IVT. Assume f is continuous. If there exists x,y\in[a,b] such that f(x)=1 and f(y)=-1, then f takes on all real values between 1 and -1 on the interval [x,y]. But this is impossible since the range is only \{-1,1\}. So f is discontinuous.
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  4. #4
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    redsox...could you elaborate on your response? specifically, this part... "then takes on all real values.... But this is impossible since the range is only ".

    I guess I don't see why f taking on all values between -1 and 1 on [x,y] means it's discontinuous if the range is from -1 to 1.
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  5. #5
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    Quote Originally Posted by spikedpunch View Post
    redsox...could you elaborate on your response? specifically, this part... "then takes on all real values.... But this is impossible since the range is only ".

    I guess I don't see why f taking on all values between -1 and 1 on [x,y] means it's discontinuous if the range is from -1 to 1.
    It is just this simple.
    How can a function be continuous if it ‘jump’ back and forth between -1~\&~1.
    That violates the intermediate value theorem.
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  6. #6
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    In the class I am in I am unable to reference the IVT, so I need to prove it some other way.

    Oh and another question. if f(a)=-1 and f(b)=1 why can't this be continuous? it could just be a straight line from (a,-1) to (b,1) and not jump around.....right?
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  7. #7
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by spikedpunch View Post
    In the class I am in I am unable to reference the IVT, so I need to prove it some other way.

    Oh and another question. if f(a)=-1 and f(b)=1 why can't this be continuous? it could just be a straight line from (a,-1) to (b,1) and not jump around.....right?
    Set \epsilon=1. \forall\delta>0, \exists x,y\in[a,b] such that f(x)=1, f(y)=-1, and |x-y|<\delta (since the range of f is only the two points \{-1,1\}, NOT the interval of values [-1,1]).

    Thus |x-y|<\delta and |f(x)-f(y)|=2>1=\epsilon, so f is discontinuous.
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  8. #8
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    I looked into your answer a bit and apparently it assumes uniform continuity, which hasn't been proven in my class and therefore cannot be used. The only tool I have to show continuity is the epsilon-delta way at a point x.

    So I'm back to square one.
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  9. #9
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    You are misreading the problem. You said "I guess I don't see why f taking on all values between -1 and 1 on [x,y] means it's discontinuous if the range is from -1 to 1."
    The range is NOT "from -1 to 1". That would be [-1,1]. The problem said that the range is {-1, 1} which is NOT an interval- it is the two numbers, -1 and 1, only.
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  10. #10
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    I understand that, thanks. But that doesn't really help or answer my question. The only definition I can use to prove discontinuity is epsilon-delta, and I can't use uniform continuity or the IVM.

    Thanks to any and all responses
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  11. #11
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    Well, then, I guess the problem depends upon what definition of "continuous" you are using. One regularly useddefinition is "f":A->B is open if and only if for every open subset, U, of B, f^{-1}(U) is an open subset of A". Here, the topology for B= {-1, 1}, since it contains of finite number of points, is the "discrete topology". Every subset is open which means that, if f is continuous, every subset of [-1, 1] must be continuous. That is not true.
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  12. #12
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    The definition of continuity we are using is: The function f is continuous at the point p=(x,y) if p is a point on f and if epsilon is any positive number, then there is a positive number delta such that if t is in the domain of f and the absolute value of t-x is less than delta, then the absolute value of f(t)-f(x) is less than epsilon.

    Redsox fan kind of used this one, but I was told by my prof that this assumes uniform continuity (?) and I'm not allowed to do so.

    Again, thanks for responses.
    Last edited by spikedpunch; November 11th 2009 at 07:21 AM.
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