# Homework Help: Real analysis, sequence of sequences convergence proof

1. Feb 27, 2015

### Perelman

1. The problem statement, all variables and given/known data
$\ell$ is the set of sequences of real numbers where only a finite number of terms is non-zero, and the distance metric is $d(x,y) = sup|x_n - y_n|$, for all n in naural-numbers
then the sequence $$u_k = {1,\frac{1}{2},\frac{1}{3},...,\frac{1}{k}, 0,0,0...}$$
and $\left\{u_k\right\}^{\infty}_{k=1}$ is what i assume to be a sequence of sequences.

I know its a Cauchy sequence. the question ask to show that {uk} is convergent or not convergent
3. The attempt at a solution

so here i just see that it will converge to the sequence $\frac{1}{k}$ from k=1 to infinity which converges to 0 but doesn't contain finite non-zero numbers. so the squenec doest belong to the set $\ell$ and its not convergent in $\ell$
Is this correct or completely wrong. I have a feeling its not and it needs more epsilons. Hard thing with proof is that there is no way to check my answer. if its correct i have some follow-up questions

2. Feb 27, 2015

### Dick

Yes, that's correct. The sequence is Cauchy but the limit isn't in $\ell$.

3. Feb 27, 2015

### Perelman

oh nice, okay the follow up. there is a set c0 that is the set of all sequences that converges to zero. then prove that $\ell$ is dense in c0.

so if im not mistaken dense means that $\ell$ must belong in c0 and all the limits of $\ell$ must also be in c0

Let $u_n$ be a sequence in $\ell$, since it only have a finite non-zero numbers then for some $n\ge N$ , $d(x_n, 0)< \epsilon$ and it have limit 0. also All the limits of $\ell$ must stay inside $c_0$, so i must show that it cant converge to a squence that doest converge to 0, here im not sure how to prove this. and i tested with the squenec that convrege to {1,1,1,1...} the same way as u_k converge to 1/k but this one is not cauchy thefore its not a limit. maybe i can show that more general in some way.

4. Feb 27, 2015

### Dick

I don't think you are trying to prove the right thing. $\ell$ is dense in $c_0$ if for every element $c$ in $c_0$ there is a sequence of elements of $\ell$ that converge to $c$.

5. Feb 27, 2015

### Perelman

but from this http://mathworld.wolfram.com/Dense.html
A set in a first-countable space is dense in if , where is the set of limit points of

so some sequence in c0 must be in $\ell$ or in the limit of $\ell$ rigth?
so by showing that all $\ell$ sequences converges to 0 or that all seq in c0 will eventually have infinite 0,0,0... elements then its in $\ell$.

but then i also must show that all the limits of $\ell$ must be in c0 like my problem above states, and im not sure how to do.. or am i thinking wrongly?

6. Feb 27, 2015

### Dick

First show every element of $c_0$ is the limit of a convergent sequence in $\ell$. That should be pretty easy. To prove every Cauchy sequence in $\ell$ converges to an element of $c_0$ prove it by contradiction. Suppose it converges to a sequence which does not converge to zero. Try to find a contradiction.

Last edited by a moderator: May 7, 2017
7. Feb 27, 2015

### Perelman

ye thats wat i was thinking and said in above post more or less. but im stuck im not that proficient in this and have tried. was thinking that the sequence that converges to some sequence that doesn't converge. like the $u_k$ and k=1 to infinity. but with just k instead of 1/k so that it will "converge" to the sequence {k,k,k,k,k.....} and the sequence that "convereges" to this is not cauchy so it cant converege to it. but this is probably not general enough. can you help me a bit further in right direction.

8. Feb 27, 2015

### Dick

Sure. A single example isn't a proof. If a sequence doesn't converge to zero what can you say about it? Carefully negate the statement that it does converge to zero. And I'll repeat, "First show every element of $c_0$ is the limit of a convergent sequence in $\ell$. That should be pretty easy." It's also the more important condition.

Last edited: Feb 27, 2015
9. Feb 28, 2015

### Perelman

well negating the statement then i get
if a sequence doesn't converge to 0 then there exist some $n>N$ such that $d(x_n, 0) > \epsilon$.
and then some contradiction happens i guess. but i cant see this.. especially if it supposed to be easy.

so i need to use the fact that its also a Cauchy sequence maybe. or that the sup of the difference of two sequences in $\ell$ will always be the max of a finite set of positive numbers.? i don't see anymore info about this set that can help me.

10. Feb 28, 2015

### Dick

How to go forward would be clearer if you had negated the statement more carefully. The negation is "there exists $\epsilon \gt 0$ such that for all $N \gt 0$ there exists $n \gt N$ such that $|x_n| > \epsilon$". In other words there exists an infinite number of terms larger than $\epsilon$. Think that would cause a problem for the Cauchy thing?

11. Feb 28, 2015

### Perelman

ok so since its a Cauchy sequence at some point you have $d(x_n, x_m) < \epsilon$ where $x_m$ is the sequence thats non-zero convergent, call it $u_k$

then the $sup|x_n, u_k|$ where $x_n = {x_0, x_1, x_2...x_i, 0,0,0...}$ with infinite zeros and $u_k = {u_0, u_1...u_k}$ for k = 1 to infinity, the difference of that is the sequence, ${d_0, d_1...d_{i}, u_{i+1}...u_{k}}$ where d is the difference $x_n - u_k$ that could be zero or some real number, and u_k that is just the difference $u_k$ and all the infinite many zeros of the $x_n$ sequence and this must be a non-zero number.

so here the sup of this can never be zero. and this shows that a sequence in $\ell$# can never converge to a non-zero-convergent sequence.
correct , or did you think of something else?

12. Feb 28, 2015

### Dick

I'm not sure I quite follow that, but the point is that since you know the infinite sequence, call it $u$, doesn't converge to zero, there are an infinite number of terms with absolute value greater than $\epsilon$. Can you see why no sequence $x$ in $\ell$ can satisfy $d(u,x)<\epsilon$?

13. Feb 28, 2015

### Perelman

no sequenec x in $\ell$ can satisfy $d(u,x) < \epsilon$ because there exist an N s.t for all $n>N , d(x_n,0) < \epsilon$, but $d(u_n,0) > \epsilon$, and then for all $n>N, |u_n - x_n| > \epsilon$ and $sup|u - x|$ must also be $> \epsilon$, making $d(x ,u) > \epsilon$ and this contradicts that it is a cauchy sequence.

14. Feb 28, 2015

### Dick

That's a bit of a muddle. Just relax and think more carefully about the quantifiers. Look at the negation of a sequence converging to zero I stated in post #10. Tell me how that implies that no sequence in $\ell$ can approximate $u$ better than $\epsilon$. Just think about it for a while. Don't worry about the Cauchy bit, I dashed that hint off running out the door this morning. You don't need it.

Too help you think about it a sequence that doesn't converge to zero is a bit like the sequence $(\epsilon, \epsilon, \epsilon, ...)$. First prove that can't be a limit of sequences in $\ell$, if that helps.

Last edited: Feb 28, 2015
15. Mar 1, 2015

### Perelman

but i think that's what i was trying to say in the last post

so that for all $n>N$ for some natural number N then

- the sequences $x$ will only now have infinite trailing zeros forever
- while $u$ will have something bigger than $\epsilon$ forever.

so when u are doing distance between those two sequences it can never be less than \epsilon since the metric takes the supremum of the difference. so that's the closest they can be.

but maybe my understanding of the infinity thing is wrong like my assumption that for all $n>N$, $x$ will just have zeros 0,0,0... forever. and that it will never have some non-zero numbers mixed in, since if those should be finite, then at some point it must stop and the zeros will continue on forever without anymore non-zero numbers. While the other sequence $u$ will still have numbers bigger than that.

is this wrong or is there is any other flaw in this, i was thinking i got it.

16. Mar 1, 2015

### Dick

You've got the right idea. I was just having a hard time interpreting your presentation. Now how about proving a sequence converging to zero is the limit of sequences in $\ell$?

17. Mar 1, 2015

### Perelman

ye it gets a bit complicated with my notation when its sequence of sequences.

well i proved that there cant be a sequence in $\ell$ that converge to a non-zero-limit sequence. so then all sequences in $\ell$ must converge to a zero-limit sequence. if im not mistaken. so then thats already proved?

also

Let u be a sequence from the set $\ell$, since it only have a finite non-zero numbers then there exist a N s.t for all $n>N$, $d(x_n,0)<\epsilon$ i.e $lim(u)=0$
this shows that all sets from $\ell$ have zero-limit.

so now all sets from $\ell$ have zero limit or a sequences in $\ell$ will converge to a zero-limit-sequence. isnt this now a proof that $\ell$ is dense in $c_o$

18. Mar 1, 2015

### Dick

You've shown that for every sequence in $\ell$ that has a limit that that limit must be in $c_0$. That doesn't show that every element of $c_0$ is a limit of a sequence in $\ell$. Pick an arbitrary element of $c_0$ and think how you might construct a sequence in $\ell$ that will converge to it. I'll repeat that I think this is the most important part of the proof. It's really what 'dense' is about.

Last edited: Mar 1, 2015
19. Mar 2, 2015

### Perelman

it seems like you know some standard way of doing these things, i have no idea how to really construct such a sequence.

20. Mar 2, 2015

### Dick

Start with a concrete example, take the infinite series (1,1/2,1/3,1/4,...) in $c_0$. Can you think of a sequence in $\ell$ that would work?

21. Mar 3, 2015

### Perelman

yea the sequence from post #1 would work for that concrete example

22. Mar 3, 2015

### Dick

Doesn't that give you some idea of what to do in general?

23. Mar 3, 2015

### Perelman

i dont get it since if im not mistaken then there is lots of more sequences that converges to zero, and the sequence in L that converges to it would probably not look like post #1 sequence at all. i might have some holes in my understanding of this so don't hesitate by giving me details that you might think i already know

24. Mar 3, 2015

### Ray Vickson

Are you sure that the 0-sequence does not belong to $\ell$? Since 0 is a finite number it is certainly true that for the 0-sequence, "all but a finite number of terms is nonzero", where here we have 0 nonzero terms. Perhaps this is defined/clarified somewhere in your textbook or notes, or perhaps you were told explicitly to not regard the 0-sequence as an element of $\ell$. That would be useful to know.

25. Mar 3, 2015

### Perelman

i cant find anything about it so maybe zero sequence could be part of it.