Why does the definition of convergence have $<\varepsilon$ instead of $\le\varepsilon$?

Question

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.)

Answer 1

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.

Answer 2

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).

Answer 3

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

Answer 4

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.