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I wanted to get more familiar with the idea of totally boundedness so I though of an example.

The space, say $X$, is the space of all bounded sequences.

The metric $d((x_n),(y_n))=\sup_{i\in{\mathbb{N}}}|x_i-y_i|$.

The "points" in $X$ are sequences such that they are bounded.

Would this be totally bounded? How would a standard argument normally go?

Sumanta
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UnsinkableSam
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  • What is your space, what are your points? There are some sharp mathematicians around but hardly anybody has a crystal ball. – Wlod AA Nov 04 '19 at 18:41
  • Of bounded sequences of what? I guess, you mean of real numbers (yes?) but it's not granted. – Wlod AA Nov 04 '19 at 18:52

2 Answers2

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Take $\epsilon=\frac{1}{2}$. We can not cover this metric space by finitely many open balls of radius $\epsilon$ as $B(x,\epsilon)\cap B(y,\epsilon)=\emptyset$ when $x=\{x_n\},y=\{y_n\}$ are two distinct sequences each having terms from the set $\{0,1\}$. Using Cantor Diagonal argument one can show, there are uncountably many sequences each having terms from the set $\{0,1\}$.

To prove the part, $B(x,\epsilon)\cap B(y,\epsilon)=\emptyset$ let, $z\in B(x,\epsilon)\cap B(y,\epsilon)$. Since $x\not= y$ we have $k\in \Bbb N$ for which $x_k\not=y_k$. Also, $x_n,y_n\in\{0,1\},\forall n\implies x_n-y_n\in\{0,\pm 1\},\forall n\implies d(x,y)=\sup_{n\in\Bbb N}|x_n-y_n|=|x_k-y_k|=1$. So, $1=d(x,y)\leq d(x,z)+d(z,y)<\epsilon+\epsilon=1,$ contradiction.

Sumanta
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  • "this metric space" -- which space? what space? – Wlod AA Nov 04 '19 at 18:53
  • Set of all bonded real or complex sequences – Sumanta Nov 04 '19 at 18:56
  • "Set of all bonded real or complex sequences" -- if that's the case, then your answer is wrong (as I've mentioned in my own answer). – Wlod AA Nov 04 '19 at 18:59
  • Ohh, you considering finite sequences also. – Sumanta Nov 04 '19 at 19:02
  • The guy who asked this question made a mistake while defining the metric. I edited this question. – Sumanta Nov 04 '19 at 19:04
  • I am sorry not to see much logic in your answer. You're taking special balls (with 0-1 center coordinates) instead of arbitrary balls. And, on the top of it, you didn't produce a point which doesn't belong to the union of these balls. ==== The whole thing is trivial but it still needs to be presented cleanly (if at all :) ). – Wlod AA Nov 04 '19 at 22:46
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The space of bounded infinite sequences of real numbers, with the metrics given by $\,\, sup\,\,$ (not by $max$ when it would not be well defined) is unbounded. Henceforth, it is not totally bounded.

Wlod AA
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