2
$\begingroup$

I'm trying to prove the following equivalence about divergence of a sequence

A real sequence $\langle x_n \rangle $ diverges

i.e. for some $\epsilon > 0$, for all $N \in \mathbb{N}$, there exists $n \geq N$ s.t. $| x_n - x | \geq \epsilon$

$\iff$ there exist infinitely many $n \in \mathbb{N}\ $ s.t. $| x_n - x | \geq \epsilon$

How should I prove the above statement?

Please help! Thank you in advance.

2 Answers 2

1

To prove $\implies$, you can use contrapositive. So given: $\forall \:\epsilon > 0,\: \exists \:N>0$ such that $\forall \:n\in \mathbb{N}$ and $n\ge N,\:|x_n-x|< \epsilon$. But then this means $\{x_n\}$ converges.

  • 0
    Is its contrapositive the following? "If there are finite n $\in$ $\mathbb{N}$ s.t. $| x_n - x | \geq \epsilon$, then a real sequence converges." – 2017-01-05
  • 0
    If $p\implies q$, then the contrapositive is $\sim q\implies \sim p$. Thats what I have written. Check it.. Your $p:$ is $x_n$ diverges and $q:$ is the remaining – 2017-01-07
  • 0
    Thanks for the comment. I know the definition of the contrapositive, but what I want to prove is $p$: "for some $\epsilon$ > 0, for all N $\in \mathbb{N}$ , there exists $n \geq N$ s.t.$ | x_n−x | \geq \epsilon$" and $q$: "there exists infinitely many $ n \in \mathbb{N}$ s.t. $| x_n - x | \geq \epsilon$" – 2017-01-08
  • 0
    anyway, I'm well aware that if I negate the definition of convergence, I can get the above statement, thanks. – 2017-01-08
0

The reverse implication $(\Leftarrow)$ isn't too hard to show. If you have infinitely many $n\in\mathbb{N}$, the set of all such $n$ has no upper bound, so for any $N\in\mathbb{N}$, there is some $n$ in that set so that $n\geq N$ and $|x_n-x|\geq\epsilon$.

In the forward direction $(\Rightarrow)$, you can construct such an infinite set by choosing $N_1$, finding $n_1\geq N_1$ with the desired property, choosing $N_2>n_1$, and so on. The set $\{n_i\}_{i=1}^\infty$ will work.