Bizarre definition of convergent sequences
I've recently begun studying Analysis I, specifically the sequence and series part. I've just come across the definition of convergence:
A sequence $(a_n)$ converges to a real number $a$ if, for every real number $\epsilon > 0$, there exists a natural number $N$ such that, for every natural number $n > N$, we have $|a_n - a| < \epsilon$.
This has the specific order quantifiers $\forall \epsilon \, \exists n \in \mathbb{N} \,\ldots$
(I've never studied formal logic before, so please do point out if the above is wrong)
I was wondering what would happen if the order of the quantifiers were reversed, that is, if we have the order $\exists \epsilon > 0 \, \forall n \in \mathbb{N} \ldots$
Then the definition would be
A sequence $(a_n)$ converges to a real number $a$ if there exists a real number $\epsilon > 0$ such that for all $N \in \mathbb{N}$ we have that $n > N$ implies $|a_n − a| < \epsilon$.
I've been thinking about this problem for a bit, and I came up with the sequence $$a_n = \frac{(-1)^n}{3} = -\frac{1}{3}, \frac{1}{3}, -\frac{1}{3}, \frac{1}{3}, \ldots$$
With our new reversed definition we could argue that the sequence $a_n$ converges to a limit $0$ by taking $\epsilon = 1$ and having $|a_n - 0| < 1$ for all natural $n$. Obviously, I know that this is a typically divergent sequence with no limit.
My question is in two parts:
- What exactly is this bizarre definition describing?
- Under what conditions would this definition give a correct proof for a convergent sequence?
1 Answer
$\begingroup$Bounded sequences
I suppose never - because 1.
To elaborate: Let $\let\epsilon\varepsilon(a_n)$ be bounded, say $|a_n|<M$. Then for $a=0$ we can let $\epsilon=M$ and see that this sequence "converges" to $0$. Or for arbitrary $a$, take $\epsilon = |a|+M$ and see that the sequence also "converges" to $a$. For the other direction assume $a_n$ "converges" to $a$ according to your definition. Pick $\epsilon >0$ accordingly. Then $|a_n-a|<\epsilon$ for all $n$, hence $(a_n)$ is bounded by $\epsilon+|a|$.
$\endgroup$ 2
