Show that the set $M = \left\{ {x \in {\ell ^1},\left| {{x_k}} \right| \le \left| {{y_k}} \right|} \right\}$ is a compact subset of $\ell^1$? Lets define the set $M = \left\{ {x \in {\ell ^1},\left| {{x_k}} \right| \le \left| {{y_k}} \right|} \right\}$ where $\ell^1$ is the set of sequences of infinite length with bounded $\ell^1$ norm, and $y$ here is fixed and in $\ell^1$. I want to show that the set $M$ is a compact subset of $\ell_1$.
Any tips are appreciated.
What I have in mind:
If I could define a continuous function, that somehow maps to $M$ or from $M$ such that what is being mapped to or from is compact, then we can conclude compactness of $M$. As for $M$ being subset, it is trivial.
 A: Let $(x_n)_n \subset M$ be a sequence. Note that $|x_n^k| \le |y^k| \le \|y\|_1$ for all $n$. Therefore, $(x_n^k)_n$ is a bounded sequence in $\mathbb R$ and by Bolzano-Weierstrass, there exists a converging subsequence $(x_{\varphi_k(n)}^k)_n$. Now let us set
$$\psi(n) = \varphi_0\circ \varphi_1 \circ \ldots \circ \varphi_n(n).$$
We have that $\psi: \mathbb N \to \mathbb N$ is strictly increasing (exercise) so that $(x_{\psi(n)})_n$ is a subsequence of $(x_n)_n$. From here it is not hard to show that $(x_{\psi(n)})_n$ converges towards some point in $M$.
A: To show that $M= \{ x \in \ell^1 \mid(\forall k\in\Bbb N):|x_k| \le |y_k| \}$ is compact, take a sequence $(x_k)_{k\in\Bbb N}$ of elements of $M$. Each $x_k$ is an element of $\ell^1$, that is, $x_k = \{x_{k,1},x_{k,2},x_{k,3},\cdots \}$.
For a fixed $j$, the sequence $(x_{k,j})_{k\in\Bbb N}$ is a bounded sequence (it is bounded by $|y_j|$). So you can choose a subsequence $(x_{k_l})_{l\in\Bbb N}$ of $(x_k)_{k\in\Bbb N}$ such that $(x_{k_l,1})_{l\in\Bbb N}$ converges. Then you can choose a subsequence $(x_{k_{l_m}})_{m\in\Bbb N}$ such that both sequence $(x_{k_{l_m},1})_{m\in\Bbb N}$ and $(x_{k_{l_m},2})_{m\in\Bbb N}$ are convergent. By using the Cantor diagonalization process, you can find a subsequence $(x_{k_n})_{n\in\Bbb N}$ of the original sequence such that each of the following limits exists$$z_j = \lim_{k\to\infty}(x_{k_{n}})_j\quad(j\in\Bbb N).$$It is clear that $(\forall k\in\Bbb N):|z_k|\leqslant|y_k|$ and therefore, if $z=(z_k)_{k\in\Bbb N}$, $z\in\ell^1$. In order to show that $\lim_{n\to\infty}x_{k_n}=z$, take $\varepsilon>0$. Choose $N$ large enough so that
$$\sum_{j=N+1}^{\infty}|y_j|< \frac\varepsilon2.$$Then choose $M$ large enough so that$$\sum_{j=1}^{M}|(x_{k_{n},j})-z_j| < \frac\varepsilon2$$for each $n\geqslant N$. Then $\left\|x_{k_{n}}-z\right\|_1<\varepsilon$ for $n\geqslant N$. Because $\varepsilon>0$ was arbitrary, then $\lim_{n\to\infty} x_{k_n}=z$ in $\ell^1$. Hence $M$ is sequentially compact and, therefore compact.
