# Sequence of bounded sequences has a subsequence with coordinatewise limit

Let $1 \leq p < \infty$ and let $r_k \geq 0$ $(k \geq 1)$ be such that $\sum_{k=1}^\infty r_k^p < \infty$. Consider $$K = \{x = (x_1, x_2, \ldots) \in \ell_p: |x_k| \leq r_k \text{ for all } k \geq 1\}.$$ Suppose that $\{x^n\}_{n=1}^{\infty} \subset K$ is a sequence in $K$. Show that there exists a subsequence $\{x^{n_k}\}_{k=1}^{\infty}$ of $\{x^n\}_{n=1}^{\infty}$ such that the coordinatewise limit $\lim_{k\to\infty} x_i^{n_k}$ exists for all $i\geq 1$.

I may have an idea, but I have no idea if it's in the right direction. First, consider $i=1$. We know that the sequence $\{x_1^n\}_{n=1}^{\infty}$ is bounded, because $|x_1^n| \leq r_1$ for all $n= 1, 2, \ldots$. So by Bolzano-Weierstrass, we know that there exists a subsequence such that $\{x_1^{n_k}\}_{n=1}^{\infty}$ has a limit. We now consider the case $i=2$ for this sequence. Using the same reasoning, we get that there exists a subsequence $\{x^{m_k}\}_{k=1}^{\infty}$ of the subsequence $\{x_1^{n_k}\}_{n=1}^{\infty}$ such that the limit $\{x_2^{m_k}\}$ has a limit. By construction, $\{x_1^{m_k}\}$ also has a limit. We can now repeat this, so we know that for every finite $N$, all $1\leq i \leq N$, the limit $\lim_{k\to\infty} x_i^{l_k}$ exists. Can I somehow expand this so this is true for all $i\geq 1$?

They said that I could use Cantor's diagonal procedure, but I'm not familiar with it in the case of subsequences. I do know the cardinality proof. I suppose these two are related?

• I think you should prove that the diagonal sequence (coming from the Cantor diagonal argument) satisfies the requirements, that $\lim\limits_{k\to\infty}x_i^{n_k}$ having in mind that each sequence constructed on the step $l$ is a subsequence of all sequences constructed before it. – Svetoslav Oct 4 '15 at 13:09
• Wow, that's genious! Thanks! – Pierre Oct 4 '15 at 13:16

$$\begin{array}{lllll} &\{x_1^n\} & \{x_2^n\} & \{x_3^n\} & \dots & \{x_l^n\} &\dots \\ &\{x_1^{n_k^1}\}\to x_1^0 & \{x_2^{n_k^1}\} & \{x_3^{n_k^1}\} & \dots & \{x_l^{n_k^1}\} &\dots \\ &\{x_1^{n_k^2}\}\to x_1^0 & \{x_2^{n_k^2}\}\to x_2^0 &\{x_3^{n_k^2}\} & \dots &\{x_l^{n_k^2}\} & \dots \\ &\{x_1^{n_k^3}\}\to x_1^0 & \{x_2^{n_k^3}\}\to x_2^0 &\{x_3^{n_k^3}\}\to x_3^0 & \dots &\{x_l^{n_k^3}\} & \dots \\ &\vdots \end{array}$$ where $n_k^1$ is the first set of subindeces, such that $x_1^{n_k^1}\xrightarrow[k \to \infty]{} x_1^0$ and $\{n_k^1\}_{k=1}^{\infty}\supset \{n_k^2\}_{k=1}^{\infty}\supset\dots$. Moreover for each $s\leq r$ we have $n_k^s\leq n_l^r,\,\,\forall k\leq l$
$$\begin{array}{lllll} & (x_1^{n_1^1} & x_2^{n_1^1} & x_3^{n_1^1} & \dots & x_l^{n_1^1} &\dots ) \\ & (x_1^{n_2^2} & x_2^{n_2^2} & x_3^{n_2^2} & \dots & x_l^{n_2^2} &\dots ) \\ & (x_1^{n_3^3} & x_2^{n_3^3} & x_3^{n_3^3} & \dots & x_l^{n_3^3} &\dots ) \\ &\vdots\\ & (x_1^{n_s^s} & x_2^{n_s^s} & x_3^{n_s^s} & \dots & x_l^{n_s^s} &\dots ) \end{array}$$. For this sequence it is true that for each fixed $i\ge 1:$ $x_i^{n_k^k}\xrightarrow[k \to \infty]{} x_i^0$