Is this the correct way to show a sequence is divergent?

[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 $x_n$?

 

[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.

 

[AxiomOfChoice]

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.

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.)

 

[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.

 

[AxiomOfChoice]

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

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?

 

[JCVD]

Right.