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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 $A \cap \{x\in l_2:|x_n| = 0, n \geq \mathbb{N}\}$.

Questions:

  • 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?
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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 $A\subset \bigcup_{j=1}^k B_\varepsilon (y^{(j)})$.

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  • $\begingroup$ 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

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