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Real Analysis limit proof problem.

  1. Oct 1, 2012 #1
    1. The problem statement, all variables and given/known data
    Define the function f:ℝ→ℝ by f(x)=0 if x is irrational and f(p/q)=1/q if p,q are integers and q>0 and the fraction is in reduced form.

    Prove f is continuous at every irrational point.


    2. Relevant equations



    3. The attempt at a solution
    We must show that lim x->a f(x)=f(a)=0 if a is irrational.
    This is clearly true for all the irrationals near a so we need to show that

    lim (p/q)->a f(x)=0
    From the limit definiton
    [itex]\forall[/itex](|p|.[itex]\epsilon[/itex]-a)>0 [itex]\exists[/itex] [itex]\delta[/itex]=(|p|.[itex]\epsilon[/itex] - a) > 0 such that |p/q - a|<[itex]\delta[/itex]

    |p/q|<[itex]\delta[/itex] + a

    |q| > [itex]\frac{|p|}{\delta + a}[/itex]

    [itex]\frac{1}{|q|}[/itex] < [itex]\frac{\delta + a}{|p|}[/itex]
    since [itex]\delta[/itex] = (|p|.[itex]\epsilon[/itex] -a)

    we simplify to

    |[itex]\frac{1}{q}[/itex]|< [itex]\epsilon[/itex]

    Is that a sufficient proof

    Any thoughts about this is appreciated.
     
  2. jcsd
  3. Oct 2, 2012 #2
    I do not follow your proof. I do not even understand what (|p|.ϵ-a) is.

    Anyway, since you have separated irrationals and rationals, then you have to prove that any sequence of rational numbers converging to an irrational has the denominator going to infinity. It looks like you are trying to prove just that, but I do not see any notion of sequences involved.
     
  4. Oct 2, 2012 #3

    HallsofIvy

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    Staff Emeritus
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    What you need is that for any interval [a, b], for a given q there exist only a finite number of p such that p/q is in that interval. Therefore, as you get closer and closer to any given point, the denominator must go to infinity.
     
  5. Oct 3, 2012 #4
    I'm trying to use the limit definition that
    lim x->a f(x)=L
    if [itex]\forall[/itex] [itex]\epsilon[/itex]'>0 [itex]\exists[/itex] [itex]\delta[/itex] > 0 such that |x-a|<[itex]\delta[/itex] implies |f(x)-L|<[itex]\epsilon[/itex]'.

    In my proof I'm trying to show that
    lim (p/q)->a f(x)=0
    [itex]\forall[/itex] [itex]\epsilon[/itex]>0 [itex]\exists[/itex] [itex]\delta[/itex] > 0 such that |[itex]\frac{p}{q}[/itex]-a|<[itex]\delta[/itex] implies |[itex]\frac{1}{q}[/itex]-0|<[itex]\epsilon[/itex].

    and to show this I've tried to use (|p|.ϵ-a) instead of [itex]\epsilon[/itex]' so that my inequality

    [itex]\frac{1}{|q|}[/itex]<[itex]\frac{\delta + a}{|p|}[/itex] simplifies into the desired result which is

    |[itex]\frac{1}{q}[/itex]|< [itex]\epsilon[/itex]
     
  6. Oct 3, 2012 #5
    That means p is both part of the ϵ specification, which is supposed to be arbitrary, and part of the value constrained by δ. That does not seem correct in any way.
     
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