# [SOLVED]Problem 15 section 5.1 from Bartle

#### bw0young0math

##### New member
hello. This is my first post.
Of course I read rules, but I may make mistakes about posting.

Now, this problem15 is in section 5.1 from Bartle.

f: (0,1)→R be bounded but such that x→0,lim f does not exist.
Show that there are two sequences (x_n), (y_n) in (0,1) with lim(x_n)=0=lim(y_n),
but such that lim f(x_n) and lim f(y_n) exist but are not equal.

I tried to understand that problem. so I tried to take an example at first.
The example is here. f(x)=sin(1/x)
If I take two sequences,x_n=1/2nπ and y_n=1/(2nπ+π/2),
lim f(x_n)=0, lim f(y_n)=1. Thus, they are not equal.

hum...

Second, I tried to solve that problem by using logic. However, I coudn't.
That problem is so complex that I couldn't change that problem to logic easily.

I want to use reduction to absurdity.
Is negation of "there are two sequences (x_n), (y_n) in (0,1) with lim(x_n)=0=lim(y_n),
but such that lim f(x_n) and lim f(y_n) exist but are not equal."
"Every sequence (x_n) with lim(x_n)=0 satisfies lim f(x_n)=L exist and they are equal." right????????? #### Opalg

##### MHB Oldtimer
Staff member
Hello bw0young0math and welcome to MHB!

The example $f(x) = \sin(1/x)$ is a good way to illustrate the truth of this result. A proof of the result is more elusive. Here's one way to reduce it to a series of logical steps.

First, take an arbitrary sequence in (0,1) that converges to zero, for example the sequence $\{1/n\}$, and look at what the function $f$ does to it. In fact, the sequence $\{f(1/n)\}$ is bounded and therefore has a convergent subsequence. So that gives you one sequence $\{x_n\}$ converging to $0$ such that $$\displaystyle \lim_{n\to\infty}f(x_n)$$ exists. Call that limit $a$.

Next, you are told that $$\displaystyle \lim_{x\to0}f(x)$$ does not exist. In particular, $$\displaystyle a\ne\lim_{x\to0}f(x)$$. Show that this means that there exists some $\varepsilon>0$ such that, for every $\delta>0$, there exists $x$ with $0<x<\delta$ such that $|f(x)-a|\geqslant\varepsilon$. (To do that, you will have to write down the negation of the $\varepsilon$-$\delta$ definition of $$\displaystyle a = \lim_{x\to0}f(x)$$.) Now apply that fact with $\delta=1/n$ to deduce the existence of a sequence $z_n\to0$ with $|f(z_n)-a|\geqslant\varepsilon$ for all $n$.

Finally, again use the fact that every bounded sequence has a convergent subsequence to get a subsequence $\{y_n\}$ of $\{z_n\}$ for which $\{f(y_n)\}$ converges to a limit different from $a$.

#### bw0young0math

##### New member
Thanks to your kinldy explanation, I understood this problem. Actually, I understood why the problem includes "f is bounded on (0,1)" Thanks.! I think that your soloution is very beautiful and good.

I thoutht of the problem by another way combining your explanation a little.

Here is my solution. Could you see and check it whether it has some troubles?

I tried to solve it by using reduction to absurdity. You can see reduction to absurdity in my solution-second part.

I have difficulties with typing. So I wrote.
^^*

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#### Opalg

##### MHB Oldtimer
Staff member
Thanks to your kinldy explanation, I understood this problem. Actually, I understood why the problem includes "f is bounded on (0,1)" Thanks.! I think that your soloution is very beautiful and good.

I thoutht of the problem by another way combining your explanation a little.

Here is my solution. Could you see and check it whether it has some troubles?

I tried to solve it by using reduction to absurdity. You can see reduction to absurdity in my solution-second part.

I have difficulties with typing. So I wrote.
^^*
The first part of the solution is fine. In the second part, you correctly get a contradiction from the assumption that $f(y_n)\to L$ for every sequence $y_n\to0.$ It follows that there must exist some sequence $y_n\to0$ such that $f(y_n)$ does not converge to $L$. But you have not proved that $f(y_n)$ converges to a limit different from $L$ (it might not converge at all). So I think that there is still some additional work to be done to complete the proof.

#### bw0young0math

##### New member
The first part of the solution is fine. In the second part, you correctly get a contradiction from the assumption that $f(y_n)\to L$ for every sequence $y_n\to0.$ It follows that there must exist some sequence $y_n\to0$ such that $f(y_n)$ does not converge to $L$. But you have not proved that $f(y_n)$ converges to a limit different from $L$ (it might not converge at all). So I think that there is still some additional work to be done to complete the proof.

I understood you meant "there must exist some sequence $y_n\to0$ such that $f(y_n)$ does not converge to $L$ "in my second part does not need to be "there exist some sequence $y_n\to0$ such that $f(y_n)$ does not converge to $M$ ≠ $L$
So my proof needs to add about M. Do I understand your intension? If it right, I think that I have to follow your way because your proof includes contents about the condition "x→0, but lim f(x) does not exist",which I didn't use in my proof.

However,
(*) there must exist some sequence $y_n\to0$ such that $f(y_n)$ does not converge to $L$

if and only if
1) there exist some sequence $y_n\to0$ such that $f(y_n)$ does not converge to $M$ ≠ $L$
or
2) there exist some sequence $y_n\to0$ such that $f(y_n)$ is divergent.

I've already proved (*) and I have to prove 1) so, I have to prove not 2).
How can I show not 2)?
For showing not 2), I have to "there is not sequence $y_n\to0$ such that $f(y_n)$ is divergent. Especially, "divergent" in the before sentence means (for example),"1,-1,1,-1,..." not "1,2,3,4,..." because f is bounded.
For showing that, the form of divegent sequence in this is special so I can't use the definition of (normal)"divergent"

Plus, I'm very very wondered why this problem is in section "continuous" function because I think definition or theorem about continuous function are not used here.
(Actually, I wrote $L$ =$f(y_n)$ , not f(0) . So I think it is not related this problem and cont..)
I just guess continuous function and [(x_n) and (f(x_n)] are related. Do you know what relation exists between cont. and this problem?

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#### Opalg

##### MHB Oldtimer
Staff member
I'm very very wondered why this problem is in section "continuous" function because I think definition or theorem about continuous function are not used here.
You are quite right, the condition that the function is continuous is not used at all, and the result would still be true if that condition were removed. All that is needed is that $f$ should be bounded and that $$\displaystyle \lim_{x\searrow0}f(x)$$ should not exist. The boundedness is used when you quote the Bolzano–Weierstrass theorem. But the real point of the problem is that it is an exercise in how to negate the convergence at $0$.

#### bw0young0math

##### New member
You are quite right, the condition that the function is continuous is not used at all, and the result would still be true if that condition were removed. All that is needed is that $f$ should be bounded and that $$\displaystyle \lim_{x\searrow0}f(x)$$ should not exist. The boundedness is used when you quote the Bolzano–Weierstrass theorem. But the real point of the problem is that it is an exercise in how to negate the convergence at $0$.
Thanks!!
I discussed this problem with my friends today.
I really appriciate you. 