2
$\begingroup$

Let $f:D\to\mathbb R$ and let $c$ be an accumulation point of $D$. Then
$(i)\lim_{x\to c}f(x)=L$ iff $(ii)$ for every sequence $(s_n)$ in $D$ that converges to $c$ with $s_n\neq c$ the sequence $(f(s_n))$ converges to $L$

I'm okay with one direction.

To prove the other direction (taking the contrapositive statement):

Suppose $L$ is not a limit of $f$ at $c$. Find a sequence $s_n$ in $D$ such that $s_n$ converges to $c$ but $(f(s_n))$ does not converge to $L$

Since $L$ is not a limit of $f$ at $c$, $\exists\epsilon>0$ such that $\forall\delta>0$ $\exists x\in D$ such that $0<|x-c|<\delta$ implies $|f(x)-L|\ge\epsilon$.

Now the book I'm reading, Steven Lay's "Analysis with an introduction to proof" goes on as follows:

" In particular, for each $n\in\mathbb N$, there exists $s_n\in D$ with
$0<|s_n-c|<1/n$ such that $|f(s_n)-L|\ge\epsilon$"

Thus exhibiting $(s_n)$ as the required sequence.

I'm not sure why is it required that $\delta$ must be related to $1/n$

. . .

ok, I want to show that there exists a sequence $s_n$ that converges to $c$ such that $(f(s_n))$ does not converge to $L$

Let $s_n$ coverge to $c$. Then $\forall \delta>0 \exists N\in \mathbb N$ such that $n\ge N \to |s_n-c|<\delta$

Now I want to make this statement into:

$\forall \delta>0 \exists s_n \in D$ such that $|s_n-c|<\delta$

please detail how that happens.

$\endgroup$
  • $\begingroup$ so what? I mean what does that do? why do I have to refer to a $\delta_n$ when I can just fix any delta and claim that the terms of the sequence $s_n$ are less then $\delta$ far away from c. I feel like I'm missing something Really obvious here.. $\endgroup$ – ak87 Feb 21 '17 at 0:11
  • $\begingroup$ The answer to your question is in effect my question. Can you please just explain to me in detail why I will need to relate delta to 1/n? I'm ok assuming that the limit of the function exists and proving that for every sequence converging in the domain of the function, the function of the sequence also converges. $\endgroup$ – ak87 Feb 21 '17 at 0:22
  • $\begingroup$ What direction are you okay? Is it $(i)\implies(ii)$? $\endgroup$ – Juniven Feb 21 '17 at 0:24
  • $\begingroup$ Okay, you are trying to use contrapositive to prove $(ii)\implies(i)$. So we need to show that $\sim (i)\implies \sim (ii)$. Yes, you are correct by assuming $\sim(i)$. Do you know how to find an equivalent statement for $\sim(ii)$? $\endgroup$ – Juniven Feb 21 '17 at 0:37
  • $\begingroup$ yes. there exists an epsilon such that for all delta 0<|x-c|< $\delta$ implies $|f(x)-L| \ge \epsilon$ $\endgroup$ – ak87 Feb 21 '17 at 0:41
1
$\begingroup$

The reason for the $1/n$ is that he is trying to produce a sequence that, by construction, converges to $c$. It would work just as well if you replaced $1/n$ with any other sequence of positive numbers that converge to $0$. Say $1/log(n)$ or $0.5^n$.

As for the argument that you gave, it is a fine proof of a different statement. Namely that if $L$ is the limit, then all sequences must converge to $L$. But it says nothing about what happens if $L$ is not the limit. (He's trying to prove that if $L$ is not the limit, then some sequence $s_n$ converges to $c$ but $f(s_n)$ does not converge to $L$.).

$\endgroup$
0
$\begingroup$

This is to see if I'm getting this straight, so please critique my answer.

To show $\lnot(ii)$, I want to find a sequence $s_n$ such that $s_n$ converges to c but $f(s_n)$ does not converge to L, given $\lnot(i)$.

$\lnot(i)$ states that $\exists\epsilon>0$ such that $\forall\delta>0$ $\exists x\in D$ such that $0<|x-c|<\delta$ implies $|f(x)-L|\ge \epsilon$

Let $s_n$ be a convergent sequence such that $\forall n \in \mathbb N$ $[s_n\in D]$. Then $\forall\delta>0$ $\exists N\in \mathbb N$ such that $n \ge N$ implies $0<|s_n-c|<\delta$ (since $\forall n \in \mathbb N$ $s_n \neq c$).

Since the existence on $x \in D$ in $\lnot (i)$ is dependent on $\delta$ I now must construct the sequence so that $\lnot (i)$ holds for all $n$, rather than just $n \ge N$, which requires $\delta$ to be related to the index of the sequence.

Thus, by $\lnot (i)$, for $\delta_n=1/n$,

$\exists s_1$ such that $0<|s_1-c|<\delta_1=1$

$\exists s_2$ such that $0<|s_2-c|<\delta_2=1/2$

$\exists s_3$ such that $0<|s_3-c|<1/3$

...

$\exists s_n$ such that $0<|s_n-c|<1/n$

Since $\forall n \in \mathbb N$ $s_n \in D$, $\lnot (i)$ applies and $0<|s_n-c|<1/n=\delta_n$ implies $|f(s_n)-L| \ge \epsilon$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.