# Showing a sequence is divergent

1. Aug 29, 2010

### AxiomOfChoice

A sequence $\{x_n\}$ in a metric space $(X,d)$ 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 $\{x_n\}$ does not converge in $(X,d)$ 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 $x\in X$ be given and show that there is some $\epsilon$ neighborhood of this point that contains at most finitely many of the $x_n$?

2. Aug 29, 2010

### JCVD

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.

3. Aug 30, 2010

### AxiomOfChoice

Ok. So my phrasing of the negation is correct, but I interpreted it incorrectly. But certainly, if I'm given a sequence $\{x_n\}$ and I want to show it doesn't converge in $(X,d)$, if I show that for every $x\in X$ there exists an $\epsilon(x) > 0$ and an $N(x)\in \mathbb N$ such that $n > N(x)$ implies $d(x,x_n) > \epsilon(x)$ , I'm good, right? (The crucial question here is whether my $N$ and my $\epsilon$ are allowed to depend on the given $x$, which I think they can.)

4. Aug 30, 2010

### JCVD

Everything in your last post is correct, but just beware that your argument will only be able to handle sequences without any convergent subsequence.

5. Aug 30, 2010

### AxiomOfChoice

Ok, thanks! But if I'm trying to show (for example) that the space $(X,d)$ 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?

6. Aug 30, 2010

Right.