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$\newcommand{\Number}[1]{\Bbb{#1}}$Let $$ \ell^2 = \left\{(x_i)_{i\in\Number{N}} \mid \text{$x_i\in\Number{R}$ for $i \in \Number{N}$; $\sum_{i=1}^\infty x_i^2<\infty$}\right\} $$ and let $$ d_2(x,y) = \left(\sum_{i=1}^\infty (x_i-y_i)^2\right)^{\frac{1}{2}};\quad x, y \in \ell^{2}. $$

How does one define (find) a bounded but not totally bounded subset?

Andrew D. Hwang
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Beverlie
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    Of course. Any ball will do. –  May 02 '17 at 13:41
  • for example, a ball which is B(0,1) in this metric, how can I understand this ball cannot have finitely constructed open cover? – Beverlie May 02 '17 at 13:44
  • Use mathjax instead of an image to show equations. It will be difficult, but for math papers, mathjax must be used. – Arbuja May 02 '17 at 13:45
  • will follow your advice later on. thank you for your link. will learn tonight – Beverlie May 02 '17 at 13:48
  • @jackerysmith A common way is using the fact that, in a metric space, a subset is totally bounded if and only if every sequence has a Cauchy subsequence. Then, you look for a sequence of vectors $v^i\in B(0,1)$ such that $\lVert v^i-v^j\rVert\ge 1/2$ for all $i,j$. There is a general lemma by Riesz which guarantees its existence for any infinitely-dimensional Banach space (see, for instance, Brézis' book), but you can make it explicit for $\ell^2$. –  May 02 '17 at 13:53
  • An equivalent definiton of totally boundedness is that each sequence within a t.b. set admits a Cauchy subsequence. For the ball $B(0,1)$, an easy counter example is $e_1,e_2,e_3,\cdots$ which are uniformly separated. (Well my comment crossed G. Sassatelli's :) – Vim May 02 '17 at 13:55

1 Answers1

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Just consider the sequences $1_i=\{0,\ldots,0,\underset{\text{$i-$th}}{1},0,\ldots\}.$ Let $S$ denote the set of all "1" sequences as I have defined above.

These are all square summable. Let $0$ denote the zero sequence in $\ell^2$. All of these sequences belong to the ball, that is $S\subset B_2(0),$ so $S$ is a bounded subset.

The set is not totally bounded because if you pick $r<1,$ then you need infinitely many open sets $B_r(x)$ where $x\in S,$ to cover $S$. Totally bounded requires that for any $r>0,$ that we can cover the set by finitely many $r$ balls. However, this is visibly not the case.

Chickenmancer
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  • any chance to make it more rigorously that for in case of r<1, we need infinitely many (open sets)? – Beverlie May 02 '17 at 13:57
  • You're welcome to do so. To see how you might want to take the distance between any of the $1_i$'s, as defined above. – Chickenmancer May 02 '17 at 13:59
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    if I describe such as below : " since for any two 1_is, their distance is root 2, we need at least more than 2 balls to cover them, thus as there exist infinitely many cases of 2 balls we shall select, we need infinitely many Br(x)" how's this description? – Beverlie May 02 '17 at 14:05
  • It seems fine for a sketch. What you can do is show that no two of these points lie in any open ball of radius less than $\sqrt{2}+\epsilon,$ and so you need at least one ball per point, and since there are infinitely many, then you have the desired claim. – Chickenmancer May 02 '17 at 14:11
  • You can also choose this answer if it satisfies your question. – Chickenmancer May 02 '17 at 14:56
  • definitely satisfactory. Could you help with another question of mine? https://math.stackexchange.com/questions/2262289/openness-of-induced-metric-space – Beverlie May 02 '17 at 15:00
  • I'm happy to help. Just ask it as a question on SE in general. Also, if my answer is satisfactory, please click accept answer." – Chickenmancer May 02 '17 at 15:02
  • how to ask SE in general? is it possible? – Beverlie May 02 '17 at 15:09
  • and how to take it as an answer? – Beverlie May 02 '17 at 15:10
  • And I meant just ask another question on Math.stackexchange so others can offer their insight in the event that mine is not enough. – Chickenmancer May 02 '17 at 15:14
  • @jackerysmith https://math.stackexchange.com/help/someone-answers – Chickenmancer May 02 '17 at 15:26
  • totally get it. Understand the notion of this community under your guidance. will follow it. – Beverlie May 03 '17 at 01:59
  • @jackerysmith I just meant click the check mark underneath the problem if it's satisfactory :p – Chickenmancer May 03 '17 at 02:02
  • Now I see. I did it! – Beverlie May 03 '17 at 03:11