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Exercise: Show that $A = \{x\in l_2:\left|x_n\right|\leq\dfrac{1}{n},n =1,2,\ldots\}$ is totally bounded, by using the fact that $\sum_{i = 1}^\infty1/n^2<\infty$ to show that $A$ is within $\epsilon$ of the set $A\cap\{x\in l_2:\left|x_n\right| =0, n\geq N\}$.

I understand that $A$ is totally bounded and I know that the answer to that question was already given here. Given $\epsilon$ we know that the interval $[-1/i,1/i]$ can be covered with a finite amount of balls with radius $\epsilon$, centered at points in the interval $[-1/i,1/i]$. Since $A\subset [-1/i,1/i]$ we know that $A$ can be covered with a finite amount of balls with radius $\epsilon$, so that $A$ is totally bounded.

What I don't understand is why $A$ is within $\epsilon$ of the set $A\cap\{x\in l_2:\left|x_n\right| =0, n\geq N\}$. If I'm not mistaken, this would mean that $$A\cap\{x\in l_2:\left|x_n\right| =0, n\geq N\} = \{\text{$x\in A$ such that the absolute value of $x_n$ is equal to $0$ for some $n\geq N$}\}$$ contains a finite set that is an $\epsilon$-net for $A$. How are we sure that there exists a point $x_0$ in this set such that $x_1,x_2\in B_\epsilon(x_0)$ for $\epsilon = 1/3$?

As far as I understand the reasoning, the answer that was given to the earlier question doesn't go into this.

Question: Why is $A$ within $\epsilon$ of the set $A\cap\{x\in l_2:\left|x_n\right| =0, n\geq N\}$?

Edit: A set $A$ in a metric space $(M,d)$ is said to be totally bounded if, given $\epsilon >0$, there exist finitely many point $x_1,\ldots,x_n\in M$ such that $A\subset \bigcup_{i =1}^n B_\epsilon (x_i)$. That is, each $x\in A$ is within $\epsilon$ of some $x_i$.

titusAdam
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  • What is meant by "within $\epsilon$"? – Jürgen Sukumaran Feb 26 '18 at 09:23
  • @TonyS.F. I edited my question with regard to your comment. – titusAdam Feb 26 '18 at 09:32
  • Yes, I know what totally bounded means. But how is one set "within $\epsilon$" of another set? Do you mean to say that $\exists$ finitely many points $x_1,\ldots,x_n\in A\cap{x\in \ell^2:|a_n|=0,n\geq N}$ such that $A\subset \bigcup\limits_{i=1}^n B_\epsilon(x_i)$? – Jürgen Sukumaran Feb 26 '18 at 09:33
  • @TonyS.F. Yes that's what I mean. I stated the exercise is exactly like it was in my book. The definition of "each $x\in A$ is within $\epsilon$ of some $x_i$" was also given like I did in the edit. – titusAdam Feb 26 '18 at 09:38
  • But you said that this would make the intersection a finite set, yet what I've written only requires it to contain a finite set. Do you see the difference? – Jürgen Sukumaran Feb 26 '18 at 09:44
  • @TonyS.F. Yes, I do. Thanks! – titusAdam Feb 26 '18 at 09:52

1 Answers1

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Let $\{x_n\} \in A$. Let $y_n =x_n$ if $|n| <N$ and 0 if $|n| \geq N$. Then $\{y_n\} $ is in $A \cap \{x \in l^{2} : x_n=0$ for $|n| \geq N\}$. And $||\{x_n\} -\{y_n\} || =\sqrt { \sum_{|n| \geq N} |x_n|^{2}} <\epsilon$ if N is large enough. Hence the distance from $\{x_n\}$ to the set $A \cap \{x \in l^{2} : x_n=0$ for $|n| \geq N\}$ is less than $\epsilon$. This is what is meant by 'within $\epsilon$'.

Kavi Rama Murthy
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  • Thank you for your answer! I don't think I fully understand what you're saying. How do you know that $|{x_n} - {y_n}| = \sqrt{\sum_{\left|n\right|\geq N}\left|x_n\right|^2} <\epsilon$? What would $|{x_n} - {y_n}|$ look like? – titusAdam Feb 28 '18 at 11:56
  • What is the definition of the norm in $l^{2}$? For the calculation of $||{x_n}-{y_n}||$ I am just using the definition. For showing that this norm is $< \epsilon $ I am using the fact that if $\sum a_n$ is a convergent series of positive numbers then $\sum_N ^{\infty} a_n \to 0$ as $N \to \infty$. – Kavi Rama Murthy Mar 01 '18 at 05:15