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Continuity of a function

  1. Sep 10, 2007 #1
    My problem is this. Let [tex]f:\mathbb{R}^{2}\longrightarrow \mathbb{R}^{2}[/tex] be a continuous function that satifies that [tex]\forall q\in\mathbb{Q}\times\mathbb{Q}[/tex] we have [tex]f(q)=q[/tex]. Proof that [tex]\forall x\in\mathbb{R}^{2}[/tex] we have [tex]f(x)=x[/tex].

    I have worked out that because it is continuous, [tex]f[/tex] satisfies that
    [tex]\forall \epsilon>0\exists\delta>0\mid \forall x\in B_{\delta}(a)\longleftrightarrow f(x)\in B_{\epsilon}(f(a))[/tex]

    and then [tex]\forall q\in\mathbb{Q}\times\mathbb{Q}[/tex] we have
    [tex]\forall \epsilon>0\exists\delta>0\mid \forall x\in B_{\delta}(q)\longleftrightarrow f(x)\in B_{\epsilon}(q)[/tex]

    therefore we have to proof that [tex]\forall x'\in\mathbb{R}^{2}[/tex] we have
    [tex]\forall \epsilon>0\exists\delta>0\mid \forall x\in B_{\delta}(x')\longleftrightarrow f(x)\in B_{\epsilon}(x')[/tex].

    It's obvious that every element of [tex]\mathbb{R}^{2}[/tex] could be approximated by some element of [tex]\mathbb{Q}\times\mathbb{Q}[/tex] or sequence in this. But, how I can link this in an expression to get what I have to proof?
     
  2. jcsd
  3. Sep 10, 2007 #2
    I'm not so sure about this but do we not know how the function maps irrational numbers, such as sqrt(2)?
     
  4. Sep 10, 2007 #3

    CompuChip

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    TimNguyen, in fact we know (they are mapped to themselves, as f is the identity map) but this is exactly what Elessar Telkont wants to show.

    Indeed you got the idea right: any real number can be approximated by a sequence of rational numbers (and therefore, pairs of reals can be approximated by pairs of rationals).
    What I would do is: Try to make this process of approximation precise (describe it in terms of epsilon-delta). Now assume what you want to prove is not true, then this should give a contradiction with the continuity (which you have also written out in epsilon-delta).

    I will take a look and post it more precisely later on (first, you give it a try yourself)
     
  5. Sep 10, 2007 #4

    CompuChip

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    Hmm, it was much easier.

    It is a familiar fact (or otherwise you should be able to easily prove it from the definition) that for continuous functions f, it holds that [tex]\lim_{n \to \infty} f(x_n) = f(\lim_{n \to \infty} x_n)[/tex] for a sequence [tex](x_n)_{n \in \mathbb{N}}[/tex].
    So describe a real pair x as the (coordinate-wise) limit of a sequence [itex]x_n[/itex] of rational pairs, then
    [tex]f(x) = f(\lim_{n \to \infty} x_n) \stackrel{*}{=} \lim_{n \to \infty} f(x_n) = \lim_{n \to \infty} x_n = x,[/tex]
    where the identity marked with a star holds by the continuity of f -- QED.

    [edit]For completeness, let me prove the claim about the limits (it's a nice exercise in epsilon-delta proofs, so you might want to try it yourself first):
    Let [itex]\epsilon > 0[/itex]. Since f is continuous, there is some [itex]\delta[/itex] such that [itex]|| x - x_n || < \delta[/itex] implies that [itex]|| f(x_n) - f(x) || < \epsilon[/itex].
    Now [itex]x_n[/itex] converging to x means that for this [itex]\delta[/itex] I can find an [itex]N[/itex] such that [itex]|| x_n - x || < \delta[/itex] as long as [itex]n > N[/itex].
    So, through the [itex]\delta[/itex] from the definition of continuity, I have found an [itex]N[/itex] for my [itex]\epsilon[/itex] such that [itex] n > N[/itex] implies [itex]|| f(x_n) - f(x) || < \epsilon [/itex], in other words,
    [tex] \lim_{n \to \infty} f(x_n) = f(x) = f( \lim_{n \to \infty} x )[/tex].
     
    Last edited: Sep 10, 2007
  6. Sep 10, 2007 #5
    Sorry about that. My math is extremely rusty since I started graduate school in physics.
     
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