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So the definition of convergence for a real sequence that I'm looking at says that $(a_{n})\to L$ means that $\forall\,\varepsilon>0\,\exists N\in\mathbb{N}:n>N\implies|a_{n}-L|<\varepsilon$.

The book I am reading says some people will say $n>N$ and some will say $n\ge N$, and it doesn't matter. I agree with that part. But what about the $<\varepsilon$? The book made no mention of that, and so I tried to figure out if $\le\varepsilon$ would work too. I feel like there must be a reason it has to be $<\varepsilon$, otherwise it would have been mentioned if they bothered to mention $n>N$ and $n\ge N$.

I thought maybe the problem would be that a lot of points would sit exactly at say, $L+\varepsilon$, but they can't sit there forever since we would be able to make $\varepsilon$ smaller and so they'd still have to get closer to $L$ anyway. Then I thought maybe the problem would be that we get an interval with only one point) like $[2,2]$, and then that would make it so convergence means the sequence's terms have to equal $L$, but since $\varepsilon$ is positive, I don't think the one-element-interval scenario can happen.

Why is $<\varepsilon$ used (or required) to accurately describe what convergence is saying about a sequence, instead of $\le\varepsilon$?

(If anyone has the time, I would greatly appreciate any explanation as to why convergence is something we want to know about a real sequence or why people want to know about limits of real sequences in general so I may be able to appreciate this definition. All I understand so far is that some infinitely long, ordered lists of real numbers get arbitrarily close to a real number and some don't. Sometimes when reading examples it does surprise me what number a sequence converges to or that a sequence doesn't actually get arbitrarily close to a number it seems to get really close to, but this is all the motivation I can think of.)

anonanon444
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  • It also doesn't matter. – Wojowu Jul 15 '18 at 11:49
  • Oh I see, thanks! I just expected the book to say something if it didn't. I have another book that used $n\ge N$ but $<\varepsilon$, so I thought the consistency in $<\varepsilon$ must have meant something important. – anonanon444 Jul 15 '18 at 11:52
  • It's equivalent, and $<$ looks nicer than $\le.$ – zhw. Jul 15 '18 at 15:07

4 Answers4

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The book made no mention of that, and so I tried to figure out if $\le\varepsilon$ would work too. I feel like there must be a reason it has to be $<\varepsilon$, otherwise it would have been mentioned if they bothered to mention $n>N$ and $n\ge N$.

All of the four notions are equivalent in the sense that a sequence converges in the sense of one of those 4 variants if and only if it does for the other ones. It's an easy exercise to verify this (and might be fun to do, so...)

However, speaking from experience, some examiners can be very picky about this sort of stuff. Hence, if you are taking an exams, I highly recommend memorizing the precise definition (even though I personally strongly disagree that this is meaningful).

Stefan Mesken
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  • Okay, thankfully, I am just self-learning so it doesn't really matter. Thanks for clearing this up! – anonanon444 Jul 15 '18 at 11:53
  • Well, wait, how come you think it's not meaningful? – anonanon444 Jul 15 '18 at 11:55
  • @anonanon444 As far as I'm concerned, among all equivalent definitions, it doesn't matter at all which one you use (since you will derive the same results). But not all examiners will let you get away with an equivalent definition -- which I find rather harmful when it comes to teaching mathematics. – Stefan Mesken Jul 15 '18 at 11:56
  • Oh, I misunderstood. I thought you meant the definition of convergence itself wasn't meaningful. I was just surprised because I just read it would be the most important definition in the the book. But thanks for taking the time to clarify! – anonanon444 Jul 15 '18 at 11:58
  • @anonanon444 Convergence is indeed key to understanding analysis. You could go as far as to say that a large part of analysis is the study of convergence. – Stefan Mesken Jul 15 '18 at 11:59
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Observe that these two definitions are equivalent in the sense that one implies the other. It's a good exercise to prove this :)

  1. $\forall\,\varepsilon>0\,\exists N\in\mathbb{N}:n>N\implies|a_{n}-L|<\varepsilon$

  2. $\forall\,\varepsilon '> 0\,\exists N\in\mathbb{N}:n>N\implies|a_{n}-L|\leq\varepsilon '$

Akira
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Let $(a_n)_{n\in\mathbb N}$ be a sequence of real numbers and let $l\in\mathbb R$. Then the assertions

  1. $(\forall\varepsilon>0)(\exists N\in\mathbb{N})(\forall n\in\mathbb{N}):n\geqslant N\implies|a_n-l|<\varepsilon$
  2. $(\forall\varepsilon>0)(\exists N\in\mathbb{N})(\forall n\in\mathbb{N}):n\geqslant N\implies|a_n-l|\leqslant\varepsilon$

are equivalent. In fact, given $\varepsilon>0$, it is clear that, if $N\in\mathbb N$ is such that$$(\forall n\in\mathbb{N}):n\geqslant N\implies|a_n-l|<\varepsilon,$$then$$(\forall n\in\mathbb{N}):n\geqslant N\implies|a_n-l|\leqslant\varepsilon\tag1$$will also hold. And if the second assertion holds and if you pick $N\in\mathbb N$ such that$$(\forall n\in\mathbb{N}):n\geqslant N\implies|a_n-l|\leqslant\frac\varepsilon2,$$then $(1)$ will also hold, since $\frac\varepsilon2<\varepsilon$.

If the assertions are equivalent, then why do we use the first one and not the second one? A matter of habit, I guess.

José Carlos Santos
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Note that $$\forall\,\varepsilon>0\,\exists N\in\mathbb{N}:n>N\implies|a_{n}-L|<\varepsilon \iff \forall\,\varepsilon>0\,\exists N\in\mathbb{N}:n>N\implies|a_{n}-L|\le \varepsilon$$

The two expressions are logically equivalent due the universal quantifier.

Thus it really does not matter.

Mohammad Riazi-Kermani
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