Compact set in lp space $1\leq p< \infty$
$$\ell_p:=\{x=(x_n):\sum_{i=1}^{\infty}|{x_i}|^p<\infty\}$$ sequance space
$$\| x\|_p:=(\sum_{i=1}^{\infty}|{x_i}|^p)^{\frac{1}{p}}$$and p norm
$$ $$
I know $ (\ell_p,\|.\|_p)$ Banach space
$$A:=\{x \in \ell_p: \forall n \in \Bbb{N}; \ |x_n| \leq n^{\frac{-2}{p}}\}$$
My question is show A is a compact subset of $\ell_p$ how can I do that I  try to show sequantial compactnes of A  but I can't please help .
 A: Consider the topological space $$X=\prod_{n=1}^\infty D_n$$
where $D_n=\{z\in\mathbb{C}: |z|\leq n^{-2/p}\}$, where $X$ is endowed with the product of the standard topology that disks have. Since this is a product of closed and bounded subsets of the plane (which are compact subsets), it is a compact space, by Tychonoff's theorem.
Define $f:X\to\ell^p$ by $f(x_n)=(x_n)$. The image of $f$ is precisely set $A$. But $f$ is continuous: Let $(x_\lambda)$ be a net converging at $x\in X$, say $x_\lambda=(x_n^\lambda)$ and $x=(x_n)$. Let $\varepsilon>0$ and we take an integer $M\geq1$ so that $\sum_{n=M+1}^\infty\frac{1}{n^2}<\frac{\varepsilon}{2^{p+1}}$, which is possible because the series converges. By the definition of the product topology, we can find $\lambda_0$ far out in the net so that for all $\lambda\geq\lambda_0$ we have that $|x_n^\lambda-x_n|<(\frac{\varepsilon}{2M})^{1/p}$ for all $n=1,\dots, M$. Therefore, for such $\lambda$ we have that
$$\|f(x_\lambda)-f(x)\|^p=\sum_n|x_n^\lambda-x_n|^p=\sum_{n=1}^M|x_n^\lambda-x_n|^p+\sum_{n=M+1}^\infty|x_n^\lambda-x_n|^p\leq$$ $$\leq\frac{\varepsilon}{2}+\sum_{n=M+1}^\infty(\frac{1}{n^{2/p}}+\frac{1}{n^{2/p}})^p\leq\frac{\varepsilon}{2}+2^p\cdot\sum_{n=M+1}^\infty\frac{1}{n^2}<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon,$$
where we used the triangle inequality together with the fact that $x^\lambda_n, x_n\in D_n$.
But a continuous function maps compact sets to compact sets, so since $X$ is compact we have that $A$ (which is equal to $f(X)$) is compact.
