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in the delta-epsilon formulation for $\lim_{n\to \infty} x_n = L$,

$\forall \epsilon >0, \exists n_0 \in \mathbb N,$

$$ |x_n-L| \leq \epsilon, \ \forall n > n_0$$

$n_0$ depends on $\epsilon$.

What else can it depend on?

Can $n_0$ depend on $x_n$?

What can it not depend on?

Thanks.

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    $\begingroup$ It can only depend on $\epsilon$. $\endgroup$
    – user84413
    Sep 24, 2014 at 19:37
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    $\begingroup$ Does it mean that $n$ can't depend on the sequence $(x_n)$? $\endgroup$
    – Tim
    Sep 24, 2014 at 19:44
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    $\begingroup$ Maybe the best way to say it is that for a given sequence $(x_n)$, $n_0$ can only depend on $\epsilon$. $\endgroup$
    – user84413
    Sep 24, 2014 at 19:48
  • $\begingroup$ Can $n_0$ depend on the sequence $(x_n)$? If yes, as a function of $x_n$ or $x_{n_0}$? $\endgroup$
    – Tim
    Sep 24, 2014 at 19:53
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    $\begingroup$ @Tim: $n_0$ should be a function of $\epsilon$ in the calculus but of course it depends of the sequence. The sequence will determine what function of $\epsilon$. $n_0=f(\epsilon)$. $\endgroup$
    – Lehs
    Sep 24, 2014 at 20:11

5 Answers 5

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$n_0=\varphi(\epsilon)$ where $\varphi$ depends on the sequence. Examples:

$x_n=\frac{1}{n}$ converges to $0$ when $n\rightarrow \infty$. Given $\epsilon>0$ chose $n_0=\varphi(\epsilon)=1+\lfloor\frac{1}{\epsilon}\rfloor>\frac{1}{\epsilon}$

$x_n=\frac{1}{n^2}$ converges to $0$ when $n\rightarrow \infty$. Given $\epsilon>0$ chose $n_0=\varphi(\epsilon)=1+\lfloor\frac{1}{\epsilon^2}\rfloor>\frac{1}{\epsilon^2}$

And so on.

The function $\varphi$ is chosen with regards to the sequence and should be a function only of $\epsilon$

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In a statement of the type $\forall \epsilon\exists \delta\forall x$, the $\delta$ may depend on anything that is written before it (as well as things you make up while defining $\delta$), but not anything written later (i.e. $\delta$ may not depend on $x$). Therefore, it is much simpler to think of the things that $\delta$ may not depend on (the things that you define after you define $\delta$). Strictly speaking, you may want to write something like

$\forall X$ metric spaces, $\forall x\in X$, $\forall (x_n)$ sequences in $X$ converging to $x$ $\forall \epsilon$ positive real numbers $\exists N\in\mathbb N$ such that $n\geq N$ implies $d(x,x_n)<\epsilon$.

When written like this, $N=N_{X,x,(x_n),\epsilon}$ may depend on all of the four previously named objects. But thinking about $N$ in this way quickly gets tedious. Note also how $x$ 'depends' on $X$, $(x_n)$ 'depends' on both $X$ and $x$, and that $n$ is not a named object at the point where $N$ is named (it is just a part of common notation for sequences). Note also that we could have included even further objects if we had wanted (say we could have considered different metrics or topologies, or different set-theoretic axioms).

In conclusion, whenever you name a new object, that object may depend on any object that you have already named. If it is important that some object does not depend on any specific previously named object, you should mention it explicitly.

Example: Consider the difference between the following three situations:

1) $\forall \epsilon>0,\exists N_{\epsilon}\in \mathbb N,\forall n\geq N_{\epsilon}: |1/n|<\epsilon$. Namely pick $N_\epsilon=\lceil 2/\epsilon\rceil$.

2) $\forall a>0,\forall \epsilon>0,\exists N_{\epsilon}\in \mathbb N,\forall n\geq N: |1/n|<\epsilon$. Namely pick $N_\epsilon=\lceil \max(a,2/\epsilon)\rceil$.

3) $\forall \epsilon>0,\exists N_{\epsilon}\in \mathbb N,\forall n\geq N: |1/n|<\epsilon$. Namely pick $N_\epsilon=\lceil \max(a,2/\epsilon)\rceil$, where $a>0$.

The only thing which matters in all three cases is that $N$ may NOT depend on $n$, because $n$ is named later than $N$.

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    $\begingroup$ +1. An example in this context is the following sentence, superficially similar to the definition of (uniform) continuity: $$\forall\varepsilon\gt0,\forall x,\forall y,\exists\delta\gt0,|x-y|\lt\delta\implies|f(x)-f(y)|\lt\varepsilon.$$ The surprise is that this holds for every function $f:\mathbb R\to\mathbb R$. $\endgroup$
    – Did
    Sep 25, 2014 at 7:58
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Consider the following sequence: $1,-1,1,-1,1,-1, \cdots$

Now, if the $n_0$ can depend on $x_n$ then this could be made to converge by looking at sub-sequences which do converge as the odd terms would converge to 1 and the even terms would converge to -1 as well but each to a different value so the overall sequence doesn't converge.

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That $$ \lim_{n\to\infty} x_n = L $$ means that for all $\epsilon >0$ there exists an $n_0$ (depending on $\epsilon$ such that if $n\geq n_0$, then $\lvert x_n -L \rvert <\epsilon$.

So $n_0$ depends only on $\epsilon$. For a given $\epsilon$ and a chosen $n_0$ it is true that for all $x_n$ for which $n\geq n_0$ you have that the distance between one of these $x_n$ and $L$ is less than $\epsilon$. The $n_0$ then "works" for all the $x_n$ with $n\geq n_0$.

So no $n_0$ doesn't depend on the the term $x_n$ in the sequence. The only way you can say that the $n_0$ depends on $x_n$ is by saying that the $n_0$ does depend on what sequence $\{x_n\}$you actually have.

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Of course you can say the choice of $n_0$ depends on the sequence $x_n$.

But you shouldn't focus on this natural number $n_0$, intuitively think of it as the existence of a cut-off point for which the tail of the sequence $x_n: n\geq n_0$ is $\epsilon$-close to the real number $L$.

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