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Homework Help: Continuity and Dense Subsets of the Real Numbers

  1. Sep 9, 2010 #1

    jgens

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    1. The problem statement, all variables and given/known data

    If [itex]f[/itex] is continuous and [itex]f(x)=0[/itex] for all [itex]x[/itex] in a dense subset of the real numbers, then [itex]f(x)=0[/itex] for all [itex]x \in \mathbb{R}[/itex].

    2. Relevant equations

    N/A

    3. The attempt at a solution

    Does this solution work? And if it does, can it be improved in some way?

    Proof: From the continuity of [itex]f[/itex], for every [itex]\varepsilon > 0[/itex] we can find a [itex]\delta > 0[/itex], such that if [itex]|x-a| < \delta[/itex], then [itex]|f(x)-f(a)| < \varepsilon[/itex]. Because [itex]f(x)=0[/itex] for all [itex]x[/itex] in a dense subset of the real numbers, it's clearly possible to choose a number [itex]x_0[/itex] from [itex](a-\delta,a+\delta)[/itex] with [itex]f(x_0)=0[/itex]. This means that [itex]|f(x_0)-f(a)| = |f(a)| < \varepsilon[/itex]. Since this is necessarily true for any given [itex]\varepsilon > 0[/itex], it follows that [itex]|f(a)| < \varepsilon[/itex] for all [itex]\varepsilon > 0[/itex]. By the Archimedean property of the real numbers, 0 is the only real number satisfying this criterion, so [itex]f(a)=0[/itex]. Clearly, this holds for any number where [itex]f[/itex] is continuous, so [itex]f(x)=0[/itex] for all [itex]x \in \mathbb{R}[/itex], completing the proof.
     
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  3. Sep 9, 2010 #2

    radou

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    Looks fine to me.
     
  4. Sep 9, 2010 #3

    jgens

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    Thanks. Is there any way that I can improve my proof then?
     
  5. Sep 9, 2010 #4

    Dick

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    Actually, your proof is assuming f is uniformly continuous, which you don't necessarily have. It's more straightforward to do a proof by contradiction. Assume f(x0)=c where c is not zero. Can you show that leads to a contradiction?
     
  6. Sep 9, 2010 #5

    jgens

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    Dick, can you explain how my proof assumes uniform continuity? And yes, I know that I can prove it that way because that was my first thought, I'm just trying to see if it's possible to do it without using a contradiction.
     
    Last edited: Sep 9, 2010
  7. Sep 9, 2010 #6

    Dick

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    Actually, I take that back. The only reason I thought that is because you mentioned delta before you mentioned 'a'. I would put an "At any point a since f is continuous at a..." at the beginning. Or at least make it clear that delta depends on a. Otherwise a careless reader like me might think you are assuming otherwise.
     
  8. Sep 9, 2010 #7

    jgens

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    Thanks! I'm sorry about that. I'll be more careful in the future.
     
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