Divergent to $\infty \Rightarrow$ Divergent?

Question

In our lecture, we defined a sequence $\left(a_n\right)_{n\in\mathbb N}$ to be divergent if it does not converge, and additionally to be divergent to $\pm \infty$, iff: $$\forall \epsilon \in \mathbb R: \exists N \in \mathbb N: \forall n > N: \pm a_n\ge \pm\epsilon$$

But I do not see how that could imply the negation of the convergence definition: $$ \neg\exists a \in \mathbb R:\forall \epsilon > 0: \exists N \in \mathbb N: \forall n \ge N: \left|a_n-a\right|<\epsilon\\ \Leftrightarrow \forall a \in \mathbb R:\exists \epsilon > 0: \forall N \in \mathbb N: \exists n \ge N: \left|a_n-a\right|>\epsilon $$

$\dots$Mainly because I just do not see how one an switch the last 2 Quantors.

How can one (elegantly) prove this implication?

Answer 1

Assuming that $$\forall \epsilon \in \mathbb R: \exists N \in \mathbb N: \forall n > N: \pm a_n\ge \pm\epsilon$$ which you will agree implies that $$\forall a \in \mathbb R: \exists N \in \mathbb N: \forall n > N: |a_n|\ge |a|+1\tag{$\ast$}$$ one is asked to show that $$\forall a \in \mathbb R:\exists \epsilon > 0: \forall N \in \mathbb N: \exists n \ge N: \left|a_n-a\right|>\epsilon\tag{$\circ$} $$ Well, I say that, by the triangular inequality, $\left|a_n-a\right|\geqslant|a_n|-|a|$ hence $(\ast)$ implies $$\forall a \in \mathbb R: \forall N \in \mathbb N: \exists n \ge N: \left|a_n-a\right|>1\tag{$\dagger$} $$ hence $(\circ)$ holds for $\epsilon=1$. Surely you can prove that $(\ast)$ implies $(\dagger)$?

Answer 2

I think I can give a proof of divergence at $ \infty \Rightarrow $ divergence: (THe other case $-\infty$ is similar.

Assume that $(a_n)$ diverges to $+\infty$. Then let $a\in \mathbb R$. Let $\epsilon = a+1 \in \mathbb R$. Then by definition there is $K\in \mathbb N$ so that $a_n \geq a+1$ for all $N\in \mathbb K$. In particular, for all $N\in \mathbb N$, there is $n\geq N$ so that

$$ a_n \geq a+1 \Rightarrow |a_n - a| \geq 1$$

So if we let $\epsilon_0 = 1/2$, then

$$\forall a\in \mathbb R, \exists \epsilon_0 = 1/2 >0, \forall N\in \mathbb N, \exists n \geq N, |a_n - a| \geq \epsilon_0.$$

Thus $(a_n)$ is divergence. Actually your negation is correct.