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?


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 $A$.

  • $\begingroup$ 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
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    $\begingroup$ 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

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