Some background before my question.
A question in my homework is as follows:
Using no negative words, say what it would mean for a sequence $\langle a_n \rangle$ to diverge.
Our definition of convergence is:
Given a sequence $\langle a_n \rangle$ we say that it converges if $\exists A \in \mathbb R$ such that $\forall \epsilon > 0, \exists N \in \mathbb N$ such that $\forall n > N$ we have $|A – a_n| < \epsilon$.
Should the definition of divergence be:
For each real number $A$, there exists an $\epsilon > 0$ such that for all $N \in \mathbb N$, there exists an $n > N$ such that $|a_n – A| \geq \epsilon$.
or would I flip each inequality?
like so:
For each real number $A$, there exists an $\epsilon \leq 0$ such that for all $N \in \mathbb N$, there exists an $n \leq N$ such that $|a_n – A| \geq \epsilon$.
Best Answer
What you want is a negation of "$\langle a_n \rangle$ converges".
This means that there is no $A \in \mathbb{R}$ such that [some other conditions]. This means that for all $A \in \mathbb{R}$ those conditions are false.
Those conditions are essentially "for all distances $\epsilon > 0$, the tail end of the sequence is $\epsilon$ away from $A$. The negation of this is that there is a distance $\epsilon$ such that no tail is $\epsilon$ away from it. Symbolically, $\exists \epsilon > 0$ such that $\forall N \in \mathbb{N}$, there is some $n > N$ such that $|A - a_n| \ge \epsilon$.
A symbolic way to look at this is knowing how to negate $\forall$ and $\exists$. Let $P$ be a proposition. We claim that $$\lnot(\forall x, P) \iff \exists x, \lnot P$$ The left hand side says "the claim that $P$ is true for all $x$ is false". The right hand side says "there is some $x$ for which $P$ is false". Similarly, $$\lnot(\exists x, P) \iff \forall x, \lnot P$$
Applying this to the definition of convergence: $$\lnot (\langle a_n \rangle \textrm{ converges}) \iff \lnot( \exists A \in \mathbb{R} \ (\forall \epsilon > 0 \ ( \exists N \in \mathbb{N} \ ( \forall n > N, \ |A - a_n| < \epsilon))))$$ $$\iff \forall A \in \mathbb{R} \ (\lnot(\forall \epsilon > 0 \ ( \exists N \in \mathbb{N} \ ( \forall n > N, \ |A - a_n| < \epsilon))))$$ $$\iff \forall A \in \mathbb{R} \ (\exists \epsilon > 0 \ (\lnot( \exists N \in \mathbb{N} \ ( \forall n > N, \ |A - a_n| < \epsilon))))$$ $$\iff \forall A \in \mathbb{R} \ (\exists \epsilon > 0 \ ( \forall N \in \mathbb{N} \ (\lnot( \forall n > N, \ |A - a_n| < \epsilon))))$$ $$\iff \forall A \in \mathbb{R} \ (\exists \epsilon > 0 \ ( \forall N \in \mathbb{N} \ ( \exists n > N, \ \lnot( |A - a_n| < \epsilon))))$$ $$\iff \forall A \in \mathbb{R} \ (\exists \epsilon > 0 \ ( \forall N \in \mathbb{N} \ ( \exists n > N, \ |A - a_n| \ge \epsilon)))$$
In words: for all possible points $A$, there is some distance $\epsilon$ such that no matter how far out $(N)$ one cuts the sequence, there is some element $(n)$ that is too far away from $A$ (i.e., $|A - a_n| \ge \epsilon$).