# Show that the set $A = \{x\in l_2: |x_n|\leq \frac{1}{n}, n = 1,2,…\}$ is closed in $l_2$.

Given are $l_2$, the collection of all real sequences for which $\sum_{n =1}^\infty |x_n|^2 < \infty$ and subset $A = \{x\in l_2: |x_n|\leq \frac{1}{n}, n = 1,2,...\}$.

I need to show that A is closed. What I've tried to do is this:

Pick a point $x \in l_2\backslash A$; there exists a $\delta$ such that $B_\delta(x) \subset l_2\backslash A \Rightarrow l_2\backslash A$ is open $\Rightarrow A$ is closed.

Question: Is my solution elaborate enough? I have just been introduced to the concept of $l_2$ and I'm not sure whether you can take balls around points as I have done above.

• $l_2 \backslash A$ is not a set you've described. For example, it also consists of such $x \in l_2$ that $x_1 > 1$, but for all other elements $|x_n| < \frac 1n$ – Elnur Oct 31 '17 at 16:02
• What is $\{x\in l_2\,:\,|x_n|>\frac1n,\ n=1,2,\ldots\}$? Is it the set of the elements of $l_2$ such that $|x_n|>\frac1n$ for some $n$? Or for every $n$? And how do you know that such a $\delta$ exists? – José Carlos Santos Oct 31 '17 at 16:04
• @Elnur what do you mean exactly? $l_2 \, A$ is not the set I've described? – titusAdam Oct 31 '17 at 16:04
• @titusAdam What is not understandable with this? $l^2\setminus A$ is not equal to $\{x : |x_n| > \tfrac 1 n,\,n=1,2,\ldots\}$. – amsmath Oct 31 '17 at 16:07
• @amsmath What would $l_2\backslash A$ be then? – titusAdam Oct 31 '17 at 16:09

Your solution is essentially circular. If for each $x\in l^2\setminus A$, there corresponds a $\delta > 0$ such that $B_\delta(x) \subset l^2\setminus A$, then $l^2\setminus A$ is open, which is equivalent to $A$ being closed. You have not proved $l^2\setminus A$ is open.
Fix $x\in l^2\setminus A$, and let $n\in \Bbb N$ such that $\lvert x_n\rvert > \frac{1}{n}$. Set $\delta = \lvert x_n\rvert - \frac{1}{n}$. Then $\delta > 0$, and $B_\delta(x) \subset l^2\setminus A$. Indeed, if $y\in B_\delta(x)$, $\lvert y_n\rvert \ge \lvert x_n\rvert - \lvert x_n - y_n\rvert > \lvert x_n\rvert - (\lvert x_n\rvert - \frac{1}{n}) = \frac{1}{n}$, so $y\in l^2\setminus A$. Since $x$ was arbitrary, $l^2\setminus A$ is open.
• Thank you for your reply! I think I get your point, however, as amsmath pointed out with his comments on my question; $l^2\backslash A$ contains sequences for which for some $n\in \mathbb{N}$, $|x_n|\ngtr \frac{1}{n}$. Shouldn't you include these sequences in your answer as well? Currently, you only look at sequences for which $n\in\mathbb{N}$ such that $|x_n|>\frac{1}{n}$, right? This is not the entire set $l_2\backslash A$ – titusAdam Oct 31 '17 at 20:04
• @titusAdam you misunderstand: I didn't say $l^2\setminus A$ is the set of all $x\in l^2$ for which $\lvert x_n\rvert > 1/n$ for all $n$, but instead the set of all $x\in l^2$ for which $\lvert x_n\rvert > 1/n$ for some $n$. – kobe Oct 31 '17 at 20:06
• But in order to show that $l_2\backslash A$ is open you need to show that for any $x\in l_2\backslash A$ we have that $B_\delta(x) \subset l_2\backslash A$ is open right? If you show that this is just the case for sequences where $n\in \mathbb{N}$ such that $|x_n| > \frac{1}{n}$, you can't conclude that $l_2\backslash A$ is open, or am I wrong? – titusAdam Oct 31 '17 at 20:10
• No. One needs to show that for every $x\in l_2\setminus A$, there corresponds a $\delta > 0$ such that $B_\delta(x)\subset l_2\setminus A$. In my argument, I let $x$ be a fixed element of $l_2\setminus A$, and found a $\delta$ that makes $B_\delta(x)\subset l_2\setminus A$. Since $x$ was arbitrarily chosen in $l^2\setminus A$, I may conclude that $l_2\setminus A$ is open. – kobe Oct 31 '17 at 20:14
• No, if $x\in l_2\setminus A$, there exists an $n$ such that $\lvert x_n\rvert > 1/n$. The $n$ is not arbitrary, it depends on $x$. – kobe Oct 31 '17 at 20:28