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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.

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    $\begingroup$ I guess that there has to be $|x_n|\leq c_n$ in the definition of $A$. $\endgroup$ – Janko Bracic Jan 28 '15 at 8:14
  • $\begingroup$ Sorry, the $|x_n|<c_n$ should be $|x_n|\leq c_n$ in the definition of $A$. Thanks for catching that. $\endgroup$ – MAM Jan 28 '15 at 14:48
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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 $x\in A$

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  • $\begingroup$ Thanks, I've changed the $|x_n|<c_n$ to $|x_n|\leq c_n$ in the definition of $A$ as it was suppose to be. $\endgroup$ – MAM Jan 28 '15 at 14:50
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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 $A$.

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  • $\begingroup$ why $c_n$ is not in $A$? $\endgroup$ – John Jan 28 '15 at 11:21
  • $\begingroup$ @John ZHANG: Because $1/n^2$ is not less than itself. This is why in the definition of A the $\lt$ sign should be replaced by the $\le$ sign $\endgroup$ – zoli Jan 28 '15 at 12:12
  • $\begingroup$ But it's less than $2/n^2$ $\endgroup$ – John Jan 28 '15 at 12:19
  • $\begingroup$ @John ZHANG: Then I misunderstand the definition of $(c_n)$. If I do then I delete my answer. I interpreted the question such that the sequence $(c_n)$ is given and it defines $A$. $\endgroup$ – zoli Jan 28 '15 at 13:33
  • $\begingroup$ Then maybe I am the person who misunderstand the question. $\endgroup$ – John Jan 28 '15 at 13:44

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