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Dense set and continuity

  1. Feb 10, 2010 #1
    This is probably very simple but I'm not sure if the last step is right.
    Let A be a dense set in the reals and f(x)=0 for all x in A. If f is continuous, prove that f(x)=0 for all x.
    Let a be in real number. By definition, for all [tex]\epsilon > 0[/tex] there exists [tex]\delta > 0[/tex] such that [tex]|x-a|< \delta \Rightarrow |f(x)-f(a)|< \epsilon [/tex].
    But in any open interval there lies element of A, so in particular: for all x in A such that
    [tex]|x-a|< \delta [/tex] we have [tex]|f(a)|< \epsilon [/tex].
    Here's where I deduce f(a)=0 and this is the step I'm not sure about. Does this make sense?
  2. jcsd
  3. Feb 10, 2010 #2
    Yes, this is correct. You are first using continuity of f, and then the fact that x can always be chosen to be in A, for any delta>0. Since epsilon>0 was arbitrary, you have shown that |f(a)| is smaller than any positive number. The only non-negative number smaller than any positive real number is 0 :-)

  4. Feb 10, 2010 #3
    Ah ok, this is what was making me unsure. I didn't know whether choosing some other epsilon and therefore having a different delta would change things. Thank you! :)
  5. Feb 10, 2010 #4
    Yeah, I was using the following:
    It can be used since your open subset around x (a such that |x-a|<delta) is always a neighbourhood of x, for any delta>0.

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