Closed set in $l_{2}$ I need to show that the set $A=\left\{ x \in l_{2} : |x_{n}| \leq \frac{1}{n}, n=1, 2, ...\right\}$ is a closed subset of $l_{2}$
I'm assuming the best way to show this is to have a sequence in A that converges to x, and show that this implies x is in A. But the manipulation is evading me.
 A: Consider continuous functionals
$$
f_n:\ell_2\to\mathbb{R}:x\mapsto x_n
$$
for each $n\in\mathbb{N}$. Note that $A_n:=(f_n)^{-1}([-1/n,1/n])$ is closed as preimage of closed set under continuous map $f_n$. Since $A=\bigcap_{n=1}^\infty A_n$, then $A$ is closed as intersection of closed sets.
A: Let $\{ x^{(m)} = (x_1^{(m)}, x_2^{(m)},\ldots ) \}_{m=1}^{\infty}$ be a sequence in $A$ and assume that $x^{(m)} \to x$ in $\ell^2$ as $m \to \infty$. To show that $A$ is closed, we need to show that $x \in A$. Write $x = (x_1,x_2,\ldots)$, then $x^{(m)} \to X$ in $\ell^2$ means that
$$
\| x^{(m)}-x \|_{\ell^2}^2 = \sum_{n=1}^{\infty} |x^{(m)}_n - x_n |^2 \to 0 \textrm{ as $m \to \infty$.}
$$
In particular, for each fixed $n$, $|x_n^{(m)} - x_n | \to 0$ as $m \to \infty$. Therefore,
$$
|x_n| \leq |x_n^{(m)} - x_n | + |x_n^{(m)}| \leq |x_n^{(m)} - x_n| + \frac{1}{n},
$$
and taking the limit as $m \to \infty$ gives that $|x_n| \leq \frac{1}{n}$. Since this holds for all $n$, we may conclude that $x \in A$, and hence that $A$ is closed.
A: Obviously 
$$
A=\left[-1,1\right]\times\left[-\frac12,\frac12\right]\times\ldots\times\left[-\frac1n,\frac1n\right]\times\ldots=\prod_{n=1}^\infty\left[-\frac1n,\frac1n\right].
$$
Given $a\in \bar{A}$, there is a sequence $(a^k)$ of elements of $A$ whose limit is $a$, i.e. $\lim_{k\to\infty}\|a-a^k\|_2=0$. Therefore, for every $k,n\ge 1$ we have:
$$
|a_n|\le |a_n^k|+|a_n-a_n^k|\le \frac1n+\|a-a^k\|_2.
$$
It follows that
$$
|a_n|\le \frac1n+\lim_{k\to\infty}\|a-a^k\|_2=\frac1n \quad \forall n\ge 1,
$$
i.e. $a=(a_n)\in A$. Hence $\bar{A}\subset A\subset \bar{A}$, i.e. $A=\bar{A}$ and $A$ is closed.
A: Hint: The complement of $A$ is 
$$
A^c=\left\{y\in l_2: |y_i|>\frac{1}{i}\text{ for some }i\right\}.
$$
If you are having trouble with your method, you could show that $A^c$ is open. 
