1
$\begingroup$

Show that the set $A = \{x \in l_2: x_n \leq \frac{1}{n}$, $n = 1,2,\ldots\}$ is compact in $l_2$. [Hint: first show that $A$ is closed. Next, use the fact that $\sum_{n=1}^{\infty} \frac{1}{n^2}$ < $\infty$ to show that $A$ is within $\epsilon$ of the set $A \cap \{x \in l_2 :|x_n| = 0, n \geq N\}$.]

Question: My main problem is that I don't know how to 'see' the set $A$. What are the actual elements of $A$? Is $A$ the collection of sequences for which each successive element gets smaller and smaller? So that we in fact have sequences $x_{i}$ where $x_{in} < \frac{1}{n}$ where $x_{in}$ is the $n_{th}$ element of the $i_{th}$ sequence?

$\endgroup$
1
$\begingroup$

I think you know why the hint is sufficient and so am only explaining why A is closed and totally bounded.

A is closed : Let $x \in l_2$ and $ \{x_n\} \in A$ s.t $ \{x_n\}$ is converging to $x$. Then each of the components $(x_n)_i$ converges to $x_i$ ( the $ith$ component of $x$) and so $|x_i| \leq \frac{1}{i}$ for all $i$ . So $x \in A$.

A is totally bounded: Fixed an $\epsilon > 0$ . Then there is $N$ s.t $\frac{1}{n} < \epsilon$ for all $n \geq N$. now $[ -\frac {1}{i} , \frac{1}{i}]$ is compact and so it can be covered by finite no. of $\epsilon$ radius balls . Let $B_i$ consists the centre of these balls which are inside $[ -\frac {1}{i} , \frac{1}{i}]$. Then consider the $\epsilon$ radius balls in $l_2$ centered at these sequences : $ \{ y \in l_2 : ( y)_n = 0$ ,for $n$ $\geq N$ and $(y)_i \in B_i$ for $i= 1(1)N \}$. Note this is a finite set and also it gives a finite $\epsilon$ net of $A$.

$\endgroup$
  • $\begingroup$ So your proof as to why $A$ is closed is: Take a random sequence in $A$ and show that it converges to a point in $A$, am I right? Also, if $|x_i| \leq \frac{1}{i}$ does this mean that all $x$ lie within the unit sphere around the origin? $\endgroup$ – Jasper Nov 1 '17 at 21:46
  • 1
    $\begingroup$ I am just using the sequential definition of a closed set of a metric space.. I.e in addition I'm assuming that the sequence converges and then showing the limit is in A....And for your next question.. A is not in unit sphere as the sequence $ \{ \frac {1}{n} \} $ is in A but its norm is pi/ √6 which is greater than 1... $\endgroup$ – Subhadip Majumder Nov 1 '17 at 22:03
  • $\begingroup$ Hmm oke, so $A$ is composed of sequences for which each successive element moves closer to 0. But as the norm of the sequence is greater than 1 we cannot say that the sequences are located in the unit sphere (even though the individual elements are)? $\endgroup$ – Jasper Nov 1 '17 at 22:28
  • $\begingroup$ @SubhadipMajumder Why are you allowed to conclude that since $[-\dfrac{1}{i}, \dfrac{1}{i}]$ is compact it can be covered by a finite no. of $\epsilon$ radius balls? $\endgroup$ – titusAdam Feb 26 '18 at 8:33
  • $\begingroup$ @titusAdam: this is because $[-\dfrac{1}{i}, \dfrac{1}{i}]$ is compact..so first cover it by balls of radius $\epsilon$ centred at every points of the interval..now what can you conclude by compactness? $\endgroup$ – Subhadip Majumder Mar 3 '18 at 11:52

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.