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Adequate proof? Spivak's Calculus ; Dense sets

  1. Jul 17, 2016 #1
    1. The problem statement, all variables and given/known data
    Let A be a dense set**. Prove that if f is continuous and f(x) = 0 for all x in A, then f(x) = 0 for all x.

    **A dense set is defined, in the book, as a set which contains a point in every open interval, such as the set of all irrational or all rational numbers.


    2. Relevant equations
    N/A


    3. The attempt at a solution
    I looked in the answer book, and spivak's proof is much different than mine. I want to know whether mine is adequate.
    I began by considering a point (α, f(α)) such that f(α)<0. I then stated that there would have to be some σ so that f(x)<0 for all x such that α-σ < x <α+σ. This, however, contradicts the initial statement that A is a dense set (since f<0 for all x in (α-σ, α+σ)).
    I had an almost identical argument for f>0, which i wish not to type. So does this prove that f(x)=0 for all x?
     
    Last edited by a moderator: Jul 17, 2016
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  3. Jul 17, 2016 #2

    LCKurtz

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    You didn't state it carefully, but what you are showing is that if f is continuous at ##\alpha## and ##f(\alpha) < 0## then you get a contradiction to A being dense. That is an indirect argument which is a reasonable way to prove it.
     
  4. Jul 17, 2016 #3
    Alright, so to correct it I should just explicitly state that f is continuous at α?
     
  5. Jul 18, 2016 #4

    LCKurtz

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    You could write: Suppose ##f## is continuous at ##\alpha## and ##f(\alpha)< 0##. Then blahdy blah blah....which is a contradiction to ##A## being dense. Therefore ##f(x)## is identically zero. But instead of assuming ##f(\alpha) < 0## I would start with ##f(\alpha) \ne 0## and use absolute value signs in your argument so you don't have to do two cases.
     
  6. Jul 18, 2016 #5
    That does seem simpler. Thanks
     
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