1
$\begingroup$

Let $A=\{(x_n)_{n\in\mathbb{N}}\;\big| \;|x_n|\leq\frac1n\quad ;n=1,2,\cdots\}$ be a subset of $\ell_2$, which is a Parallelepiped in $\ell_2$.

I think I've stuck in a problem with notation !

Assume $\left\{\begin{array}{l} x^{(1)}:=\left(x_1^{(1)},x_2^{(1)},\cdots\right)\\ x^{(2)}:=\left(x_1^{(2)},x_2^{(2)},\cdots\right)\\ \quad\qquad\vdots \end{array}\right.$ is a given sequence in $A$.

Since $[-1,1]$ is compact, for the sequence $\{x_1^{(n)}\}_{n\in\mathbb N}$,there's a subsequence $\{x_1^{(f_1(n))}\}_{n\in\mathbb N}$ which converges to some $\alpha_1$.

Since $[-\frac12,\frac12]$ is compact, for the sequence $\{x_2^{(f_1(n))}\}_{n\in\mathbb N}$,there's a subsequence $\{x_2^{(f_2(f_1(n)))}\}_{n\in\mathbb N}$ which converges to some $\alpha_2$.

Similary we'll have nested subsequences of indexes, i.e $\left\{f_n o\cdots of_1(n)\right\}_{n\in\mathbb N}$.

Now I want to find my desired subsequence $\{x^{(n_k)}\}_{k\in\mathbb N}$ which tends to $(\alpha_1,\alpha_2,\cdots)$ in each component . Then I'll can prove the convergence is still validate in $\ell_2$.

But I'm really confused with my notation for subsequences.

Can anyone derive the convergent subsequence of $\{x^{(n)}\}_{n\in\mathbb N}$ to $(\alpha_1,\alpha_2,\cdots)$ and completes the proof of compactness ?


EDIT

Proving totally Boundedness of $A$ is an easy task and completes the proof. But I want to overcome the fear of working with nested subsequences which is essential in several proofs of compactness !

$\endgroup$
  • $\begingroup$ It is a good idea to denoe a subsequence of $x^{(n)}$ by $y^{(n)}=x^{(f_1(n))}$ where $f_1:\mathbb N\to\mathbb N$ is strictly increasing. But then a subsequence of $y$ is of the form $ y^{(f_2(n))}=x^{(f_1\circ f_2 (n))}$. $\endgroup$ – Jochen Oct 15 '15 at 8:09
  • $\begingroup$ And then ? What is the convergent subsequence in this manner to complete the proof ? $\endgroup$ – Fardad Pouran Oct 15 '15 at 10:50
  • 1
    $\begingroup$ $f(n)=f_1\circ\cdots\circ f_n(n)$ is then a subsequence of each $f_k$. $\endgroup$ – Jochen Oct 16 '15 at 6:20
1
$\begingroup$

There are many notations, which fulfill the task being somehow readable, I will present the one I liked most, when I encoutered it, to my pity, I forgot where: We will count.

So $(x^{(n)})_n$ is a sequence in $A$, and hence $(x_1^{(n)})$ has a convergent subsequence, let the corresponding subsequence of $(x^{(n)})$ be denoted by $(\def\no#1{{}^{#1}x^{(n)}}\no 1)$, as it is the first chosen subsequence. Going on inductively: Suppose for $j < k$ we have found nested subsequences $(\no j)$ such that for $i \le j$, $(\no j_i)$ converges. As $(\no {k-1}_k)_n$ is bounded, it has a convergent subsequence, the corresponding subsequence of $(\no{k-1})$ we will denote $(\no k)$. By this we have constructed sequences $(\no k)$ such that:

  1. For each $k$, $(\no{k-1})$ is a subsequence of $(\no k)$.
  2. For each $j \le k$, $(\no k_j)$ converges in $\mathbf R$, to $x_j$, say.

To find the convergent subsequence now, we will use the "diagonal". Think of $(\no k)$ be written in lines, one below the other: $$ \begin{matrix} \no 1_1 & \no 1_2 & \no 1_3 & \cdots \\ \no 2_1 & \no 2_2 & \no 2_3 & \cdots \\ \no 3_1 & \no 3_2 & \no 3_3 & \cdots \\ \vdots & \vdots & \vdots \end{matrix} $$ Now define $y^{(n)} := \no n$, that is the $n$-th term of $y$ is the $n$-th term of the $n$-th choosen subsequence, the $y^{(n)}$ form the diagonal of the diagram above. By this trick, we have that $(y^{(n)})_{n\ge k}$ (we cut before the $k$-th term) is a subsequence of $(\no k)$ (by property 1. above), and therefore, by 2. above $y^{(n)}_k \to x_k$. As $k$ was arbitrary, we have that $(y^{(n)})_n$ converges to $y := (x_k)$ in all coordinates. Now continue and prove that $y^{(n)} \to y$ in $\ell^2$.

$\endgroup$
  • $\begingroup$ Great. I exactly needed such a diagonal selection of something that was hidden to me. $\endgroup$ – Fardad Pouran Oct 15 '15 at 11:52

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.