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:

  1. What exactly is this bizarre definition describing?
  2. Under what conditions would this definition give a correct proof for a convergent sequence?
  • $\begingroup$ "For all $\epsilon >0$ there exists an $N$..." means that for arbitrarily small quantities, there is a point in the sequence such that all terms after are within the small quantity of the limit $a$. "There exists an $\epsilon >0$ such that for all $N$..." only means that it is bounded (and that $a$ carries no immediate significance to the sequence). $\endgroup$ – Taylor Jul 4 '15 at 14:19
  • $\begingroup$ Related: Slightly changing the formal definition of continuity. Particularly, my answer discusses the four conditions obtained by permuting the order of the quantifiers "for every" and "there exists" and which variable ($\epsilon$ or $\delta$, cf. $N$) each quantifies. $\endgroup$ – Andrew D. Hwang Jul 4 '15 at 14:35
  1. Bounded sequences

  2. 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 $\epsilon+|a|$.

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  • $\begingroup$ I see, what's the distinction between a bounded and a convergent sequence. Also, I've heard often that a bounded sequence always attains its bounds, what does that mean? $\endgroup$ – Zain Patel Jul 4 '15 at 14:05
  • 3
    $\begingroup$ @ZainPatel If attaining means, some $a_{n_1}$ is equal the lower bound and some $a_{n_2}$ is equal the upper bound, then it isn't true. However, a bounded sequence does attain it's infimum/supremum or it has subsequences which converge to them. $\endgroup$ – user251257 Jul 4 '15 at 14:14

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