Proving a subset of $l_2$ is closed

Question

Let $l_2$ be the set of all real sequences $x=(x_n)$ such that $\sum|{x_n}|^2 <\infty$ and define the norm $||x_n||_2=(\sum\limits_{n=1}^{\infty}|x_n|)^{\frac{1}{2}}$. I want to show that $A=\{ x\in l_2 : |x_n|\leq c_n, n=1,2,... \}$ where $c_n\geq 0$ for all $n$ is closed in $l_2$. I've tried, with out much luck, using the sequential definition of closed, that is if $(x_n^i)\subset A$ and $(x_n^i)$ converges to some $(x_n)$ in $l_2$ then $(x_n)$ is in $A$. Any hints or suggestions on how to go about showing that $A$ is closed would be greatly appreciated.

Answer 1

If the sequence $(c_n)$ is fixed for $A$, then $<$ should be changed to $\le$ as suggested by Janko. Please see Zoli's answer for a counterexample.

If $(x_k^n)\in A$ and $(x_k^n)$ converges to some $(x_k)$ in $l_2$ as $n\to \infty$ then we show $(x_k)$ is in $A$.

Since we have $(x_k^n)$ converges to $(x_k)$ in $l_2$, hence $|x_k^n-x_k|\le\sum|x_k^n-x_k|^2\to 0$ as $n\to \infty,\forall k$, thus we know $(x_k^n)\to x_k$ as $n\to \infty$ for all $k$. Since $(x_k^n)\in A$, $|x_k^n|\le c_k$, pass $n\to \infty$, we have $|x_k|\le c_k$

Hence $x\in A$

Answer 2

A counterexample for the "sequential proof" (If $\lt$ is changed to $\le$ in the definition of $A$ as suggested by Janko above then my counter example does not work.):

Take $(c_n)=(\frac{1}{n^2})$ and $(x_n^i)=\big(\frac{1}{(n+1/i)^2}\big)$.

The $(x_n^i$)'s are in $A$ defined by ($c_n$). $(c_n)$ itself is not in $A$ but it is in $l_2$.

$(x_n^i)$ converges to $(c_n)$, a series not in $A$.