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.

Thanks in advance!

  • 1
    $\begingroup$ $l_2 \backslash A$ 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 $|x_n| < \frac 1n$ $\endgroup$ – Elnur Oct 31 '17 at 16:02
  • $\begingroup$ 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? $\endgroup$ – José Carlos Santos Oct 31 '17 at 16:04
  • $\begingroup$ @Elnur what do you mean exactly? $l_2 \, A $ is not the set I've described? $\endgroup$ – titusAdam Oct 31 '17 at 16:04
  • $\begingroup$ @titusAdam What is not understandable with this? $l^2\setminus A$ is not equal to $\{x : |x_n| > \tfrac 1 n,\,n=1,2,\ldots\}$. $\endgroup$ – amsmath Oct 31 '17 at 16:07
  • $\begingroup$ @amsmath What would $l_2\backslash A$ be then? $\endgroup$ – 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.

  • $\begingroup$ 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 $l_2\backslash A$ $\endgroup$ – titusAdam Oct 31 '17 at 20:04
  • $\begingroup$ @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 $n$. $\endgroup$ – kobe Oct 31 '17 at 20:06
  • $\begingroup$ 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? $\endgroup$ – titusAdam Oct 31 '17 at 20:10
  • $\begingroup$ 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. $\endgroup$ – kobe Oct 31 '17 at 20:14
  • 1
    $\begingroup$ 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 $x$. $\endgroup$ – kobe Oct 31 '17 at 20:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.