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Showing a sequence is divergent

  1. Aug 29, 2010 #1
    A sequence [itex]\{x_n\}[/itex] in a metric space [itex](X,d)[/itex] converges iff

    (\exists x\in X)(\forall \epsilon > 0)(\exists N \in \mathbb N)(\forall n > N)(d(x_n,x) < \epsilon).

    Am I correct when I assert that the negation of this is: A sequence [itex]\{x_n\}[/itex] does not converge in [itex](X,d)[/itex] iff

    (\forall x\in X)(\exists \epsilon > 0)(\forall N \in \mathbb N)(\exists n > N)(d(x_n,x) \geq \epsilon)?

    So, if I'm trying to show a sequence does not converge, I let [itex]x\in X[/itex] be given and show that there is some [itex]\epsilon[/itex] neighborhood of this point that contains at most finitely many of the [itex]x_n[/itex]?
  2. jcsd
  3. Aug 29, 2010 #2
    Your original assertion of negation is correct, but your final statement is incorrect: for a given x, you must only show there is a neighborhood of x outside of which infinitely many x_n lie. The sequence 0,1,0,1,0,1,0,1,... does not converge to 0, but any neighborhood of 0 contains infinitely many x_n.
  4. Aug 30, 2010 #3
    Ok. So my phrasing of the negation is correct, but I interpreted it incorrectly. But certainly, if I'm given a sequence [itex]\{x_n\}[/itex] and I want to show it doesn't converge in [itex](X,d)[/itex], if I show that for every [itex]x\in X[/itex] there exists an [itex]\epsilon(x) > 0[/itex] and an [itex]N(x)\in \mathbb N[/itex] such that [itex]n > N(x)[/itex] implies [itex]d(x,x_n) > \epsilon(x)[/itex] , I'm good, right? (The crucial question here is whether my [itex]N[/itex] and my [itex]\epsilon[/itex] are allowed to depend on the given [itex]x[/itex], which I think they can.)
  5. Aug 30, 2010 #4
    Everything in your last post is correct, but just beware that your argument will only be able to handle sequences without any convergent subsequence.
  6. Aug 30, 2010 #5
    Ok, thanks! But if I'm trying to show (for example) that the space [itex](X,d)[/itex] is not complete, then I'd *need* an argument like the above, right? Because if my sequence is Cauchy and has a convergent subsequence, the whole sequence converges to the limit of the subsequence, right?
  7. Aug 30, 2010 #6
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