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 ?


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 !

  • $\begingroup$ 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

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 $\ell^2$.

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

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