Let $A = \{x\in l_2:|x_n|\leq \frac{1}{n}, n = 1,2,...\}$ be a subset of $l_2$, where $l_2$ is the collection of real sequences for which $\sum\limits_{n = 1}^{\infty}|x_n|^2 <\infty$.

Exercise: Show that $A$ is totally bounded.

I know that $A$ is totally bounded if for every $\epsilon > 0$, there exists a finite number of points $x_1, ..., x_n \in l_2$, such that $A \subset \bigcup_{i = 1}^{n}B_\epsilon(x_i)$. That is, each $x \in A$ is within some $\epsilon$ of some $x_i$.

I'm given the following hint: use the fact that $\sum\limits_{n = 1}^{\infty}\frac{1}{n^2}<\infty$ to show that $A$ is within some $\epsilon$ of the set $A \cap \{x\in l_2:|x_n| = 0, n \geq \mathbb{N}\}$.


  • How would you use the hint to show that $A$ is within some $\epsilon$ of the set $A \cap \{x\in l_2:|x_n| = 0, n \geq \mathbb{N}\}$?
  • Why would the fact that $A$ is within some $\epsilon$ of the set $A\cap \{x\in l_2:|x_n| = 0, n\geq \mathbb{N}\}$ imply that $A$ is totally bounded?

The idea is the following. Given $\varepsilon > 0$, you can find $N\in\mathbb{N}$ such that $$ \sum_{n=N+1}^\infty \frac{1}{n^2} < \frac{\varepsilon^2}{4}. $$ Let $$ A_N := \{x \in A:\ x_n = 0\ \forall n > N\}. $$ It is easily seen that $A_N$ is compact, so it can be covered by a finite number of balls $B_j \equiv B_{\varepsilon /2 }(y^{(j)})$, $j=1,\ldots, k$.

Given $x = (x_1, x_2, \ldots)\in A$, we have that $x^{(N)} := (x_1, \ldots, x_N, 0, 0, \ldots)$ lies in some ball $B_j$. Since $\|x - x^{(N)}\| < \varepsilon / 2$, we have that $\|x - y^{(j)}\| < \varepsilon$, hence $A\subset \bigcup_{j=1}^k B_\varepsilon (y^{(j)})$.

  • $\begingroup$ It's not that clear to me that any point $x\in A\backslash A_N$ lies in the ball $B_\epsilon(0)$ of $l_2$! Pick the sequence $x_n =\frac{1}{2n}$ and choose $\epsilon < \frac{1}{3}$, then $x_n\notin B_\epsilon(0)$ right? $\endgroup$ – titusAdam Nov 1 '17 at 11:56
  • $\begingroup$ Yes, you are right. I have edited my answer. $\endgroup$ – Rigel Nov 1 '17 at 12:35
  • $\begingroup$ how is $A_N $ compact. Is it because $A_N$ is closed in $\ell2$ which is complete. Also where does $y^{(j)}$ lie $\endgroup$ – Abhay Oct 29 '19 at 0:33

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.