Real analysis, sequence of sequences convergence proof

[Perelman]

Hmm. Ok, $a=(a_1,a_2,a_3,...)$, a sequence that converges to 0. How about if you define $u_k=(a_1,a_2,...,a_k,0,0,0,...)$? What is the limit of the sequence $u_k$?

it will converges to $a$, i think i found some kind of proof now.

let $a = (a_1,a_2,a_3,a_k..)$ be a zeroconvergent sequence.
then for all $n,m > N , d(a_n, a_m)<\epsilon$, for some $N$ then
let $n$ be $n+1$ and $m$ be $n+a$ s.t $d(a_{n+1}, a_{n+a})<\epsilon.$

then the sequence of sequences $L = (a_1,a_2,...,a_k,0,0,0,...)$ for k=1 to infinity in $\ell$ would converge to $a$
because $d(L_n , L_{n+a}) = sup|L_n - L_{n+a}|$ which all the elements up to and including $L_n$ turns zero and you end up with the supremum of the difference, $L_{n+1}$ must be the supremum, since its a decreasing sequence.
since $L_{n+1}$ is the $a_{n+1}$ of the sequence $a$, and $L_{n+a} = a_{n+a}$ then
$d(L_{n+1} , L_{n+a}) < \epsilon$. for all $n>N$

 

[Dick]

it will converges to $a$, i think i found some kind of proof now.

let $a = (a_1,a_2,a_3,a_k..)$ be a zeroconvergent sequence.
then for all $n,m > N , d(a_n, a_m)<\epsilon$, for some $N$ then
let $n$ be $n+1$ and $m$ be $n+a$ s.t $d(a_{n+1}, a_{n+a})<\epsilon.$

then the sequence of sequences $L = (a_1,a_2,...,a_k,0,0,0,...)$ for k=1 to infinity in $\ell$ would converge to $a$
because $d(L_n , L_{n+a}) = sup|L_n - L_{n+a}|$ which all the elements up to and including $L_n$ turns zero and you end up with the supremum of the difference, $L_{n+1}$ must be the supremum, since its a decreasing sequence.
since $L_{n+1}$ is the $a_{n+1}$ of the sequence $a$, and $L_{n+a} = a_{n+a}$ then
$d(L_{n+1} , L_{n+a}) < \epsilon$. for all $n>N$

It doesn't say that the sequence $a$ is decreasing, it just says that it converges to zero. I think you've got the right idea but your notation is pretty confusing. If you want to set $L_k = (a_1,a_2,...,a_k,0,0,0,...)$ then you want to prove that the sequence $(L_1,L_2,L_3,...)$ converges to $a$. So for all $\epsilon \gt 0$ there exists an $N$ such that for all $n \gt N$ that $d(L_n,a) \lt \epsilon$. Can you try to state a proof of that a bit more clearly?

 

[Perelman]

It doesn't say that the sequence $a$ is decreasing, it just says that it converges to zero. I think you've got the right idea but your notation is pretty confusing. If you want to set $L_k = (a_1,a_2,...,a_k,0,0,0,...)$ then you want to prove that the sequence $(L_1,L_2,L_3,...)$ converges to $a$. So for all $\epsilon \gt 0$ there exists an $N$ such that for all $n \gt N$ that $d(L_n,a) \lt \epsilon$. Can you try to state a proof of that a bit more clearly?

let $a = (a_1,a_2,a_3,a_k..)$ be a zero-convergent sequence.
then the sequence of sequences $L_n = (a_1,a_2,...,a_n,0,0,0,...)$ for n=1 to infinity in $\ell$ would converge to $a$
because $d(L_n , a) = sup|L_n - a|$, all the elements up to and including $a_n$ turns to zero, therfor $d(L_n , a) = sup|(a_{n+1}, a_{n+2}...)|$
since for all $n>N, d(a_n,0) < \epsilon$ for some N then
$d(L_n , a) = sup|(a_{n+1}, a_{n+2}...)| < \epsilon$ , for all $n>N$ for some N

 

[Dick]

let $a = (a_1,a_2,a_3,a_k..)$ be a zero-convergent sequence.
then the sequence of sequences $L_n = (a_1,a_2,...,a_n,0,0,0,...)$ for n=1 to infinity in $\ell$ would converge to $a$
because $d(L_n , a) = sup|L_n - a|$, all the elements up to and including $a_n$ turns to zero, therfor $d(L_n , a) = sup|(a_{n+1}, a_{n+2}...)|$
since for all $n>N, d(a_n,0) < \epsilon$ for some N then
$d(L_n , a) = sup|(a_{n+1}, a_{n+2}...)| < \epsilon$ , for all $n>N$ for some N

Better. You should mention why there exists such an N. It's because $a$ converges to zero, right? Spelling out more reasons makes for a more readable proof.

 

[Perelman]

Better. You should mention why there exists such an N. It's because $a$ converges to zero, right? Spelling out more reasons makes for a more readable proof.

okay ill add that in in my paper. but is this the whole proof or do i also need the other things? since now $c_0$ could just be a boundary around $\ell$ but i guess that's also counted as $\ell$ being dense. maybe i also need to add proof that $\ell \subset c_0$

 

[Dick]

okay ill add that in in my paper. but is this the whole proof or do i also need the other things? since now $c_0$ could just be a boundary around $\ell$ but i guess that's also counted as $\ell$ being dense. maybe i also need to add proof that $\ell \subset c_0$

I think you've done all you need to do. But if you aren't convinced of that you should do more to convince yourself.

 

[Perelman]

I think you've done all you need to do. But if you aren't convinced of that you should do more to convince yourself.

but i mean we did three things

everything from $\ell$ is in $c_0$
all seqeunce in $\ell$ converger in $c_0$
and all "points" in $c_0$ has a seqeunce in $\ell$ that converge to it

so its all those together right?. also thanks a lot for your help :)

 

[Dick]

Sort of, we didn't prove all sequences in $\ell$ converge. We proved that if a sequence in $\ell$ converges, then it converges to a limit in $c_0$. You're welcome!