# About the epsilon definition of a convergent sequence. Is this definition equivalent?

I read that it is appreciated to include the context and motivation of a question. I may have overdone this a little bit in this question. To summarize, my question is: Are the two blockquoted definitions equivalent ?

I have a little obsession with the following definition in real analysis:

A sequences $(s_n)$ of real numbers is said to converge to the real number $L$ provided that

$$\text{for each } \epsilon>0 \text{ there exists a number N such that n>N implies }|s_n-L|<\epsilon.$$

I'm not sure why. Maybe this is because, this is the first mathematical definition that I've learned at university. Maybe it is because I always thought that there is nothing that I couldn't understand, and reading this defintion made me doubt my own abilities.

Or maybe it is because, that allthough it is more than 1 years ago since I first read this definition, and I can write it down for you at any moment, I can apply it for you at any moment, but it always takes me some minutes to recapture why this definition makes sense. In other words, why this definition is equivalent with the intuive idea of a convergent sequence.

So every now and then, I try to go to this thought process. I pretend that I've never seen this definition before, and I try to indepedenlty come to conclussion that this sequence does correspond with the intuive idea of convergent sequence.

The very first time I did this thought process, it took me hours, honestly. But even after a year, it takes me several minutes. This fact frustrated me a little bit, and I began to analyse why it takes my brain so long, before he gets it. This is my conclussion:

1. The number epsilon get introdcued as a postive number right in the beginnning, but it gets only some real meaning at the very last.
2. The part $\forall n>N$. For some reason, it takes me time before I get to uderstand that this just means, let's restrict the domain of $s_n$ to $n>N$.
3. The phrase $|s_n-L|<\epsilon$. It takes me some time before I see that this means $s_n \in (L-\epsilon,L+\epsilon)$.
4. The nature of this logical expression is just complicated: $\forall ... \exists ... \forall ... :$

So, as I've analysed this so intensly now, I don't think it will ever cost me more than one minute to go through my thought process (of indepedenlty understanding that the definition corresponds with the intuive idea of a convergent sequence).

But I am also thinking, could I rephrase this definition so that it is more intuive ? Well, it hard to judge that, but at least I've tried, and I am especially curious if this definition is equivalent with the standard one:

A sequence $(s_n)$ converges to $L$ if there exists a function $f:\mathbb{R}_{>0}\to\mathbb{R}$ such that

$$\forall x >0: (L-x,L+x)\supset \{s_n : n>f(x)\}$$

or using more common notation:

A sequence $(s_n)$ converges to $L$ if there exists a function $N:\mathbb{R}_{>0}\to\mathbb{R}:\epsilon \mapsto N(\epsilon)$ such that

$$\forall \epsilon >0:(L-\epsilon,L+\epsilon)\supset \{s_n : n>N(\epsilon)\}$$

• Btw, I'm not sure what the standard way is for writing the image set of a sequence. I also thought about writing $s_{n>N(\epsilon)}$ or $s(\{n\in \mathbb{N}: n>N(\epsilon)\})$. – Kasper Mar 14 '14 at 15:00
• I think that the second one (that with the function $N$) is equivalent to the standard definition. – Siminore Mar 14 '14 at 15:21
• I maybe missing some tiny logical bit but I think your definitions are fine and equivalent to the usual one. What I can't understand is why would anyone think these definitions are, in some more or less mundane sense, "more intuitive" than the usual one...you could also "word out" the definition: The limit of $\;\{s_n\}\;$ is a number $\;L\;$ iff for any $\;\epsilon\;$ neighbourhood of $\;L\;$, all but a finite number of elements of the sequence are within that neighbourhood. – DonAntonio Mar 14 '14 at 15:27
• @Donantonio If I see this definition, I immediately see the intuitive picture, of a convergent sequence. With the normal definition, I need to think hard to get that picture. But this probably very individual different :-) You definition is hard for me, as I don't know what a neigbourhood is :P – Kasper Mar 14 '14 at 15:34
• $(L-\epsilon,L+\epsilon)$ is an $\epsilon$-neighborhood of $L$. – John Habert Mar 14 '14 at 15:36

Let $x>0$. By the normal definition, there exists a $N$ such that $n>N$ implies $|s_n-L|<x$. Define $f(x)=N$. We can repeat this step for every arbitrary $x>0$, so in this way we construct a function $f:\mathbb{R}_{>0}\to\mathbb{R}$ such that for every $x>0$ we have that $n>f(x)$ implies that $|s_n-L|<x$. We are now going to prove that $(L-x,L+x)\supset \{s_n : n>f(x)\}$. Let $n>f(x)$. Then $|s_n-L|<x$, so $s_n \in (L-x,L+x)$.
Let $\epsilon >0$. By your definition, there exist a function $N:\mathbb{R}_{>0}\to\mathbb{R}:\epsilon \mapsto N(\epsilon)$ such that $$(L-\epsilon,L+\epsilon)\supset \{s_n : n>N(\epsilon)\}$$
Choose $N=N(\epsilon)$. Then if $n>N$ then $s_n \in (L-\epsilon,L+\epsilon)$. Therefore $n>N$ implies $|s_n-L|<\epsilon$.