If we look a set $\mathbb{N}$ in metric space $(\mathbb{R},d_2)$ is $\mathbb{N}$ there a totally bounded space? What about looking set $\mathbb{N}$ in metric space $(\mathbb{R},d)$ where $d$ is discrete metric?
Definition of totally bounded space:
A metric space $(M,d)$ is totally bounded if and only if for every real number $\epsilon>0$, there exists a finite collection of open balls in M of radius $\epsilon$ whose union contains M.
My work
It is pretty clear that is not the case but I need to show this by using definition of totally bounded space. I first start looking $B(n,\frac{1}{2})$ ball around every natural number, but $\mathbb{N}$ is not a finite set so we don't have a finite cover of set I think that this could be a problem?
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josf
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What is $(\mathbb R,d_2)$? – Kavi Rama Murthy Oct 15 '20 at 07:13
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$d_2$ is normal Euclidian metric. – josf Oct 15 '20 at 07:14
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Why don't you show us your definition of a totally bounded space ? – Oct 15 '20 at 07:18
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You cannot cover $\mathbb N$ by a finite number of open balls of raidius $\frac 1 2$ in either of the two metrics. – Kavi Rama Murthy Oct 15 '20 at 07:19
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What is your question precisely ? – Oct 15 '20 at 07:22
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@YvesDaoust is N totally bounded space in this two metric space and why? And I add my definition – josf Oct 15 '20 at 07:23
1 Answers
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Hint:
Let any finite cover $C=\cup_{i=1}^n B(b_n,\epsilon)$. The number
$$\max_{1\le i\le n}(b_n+\epsilon)$$ is an upper bound to the integers contained in $C$ and cannot contain
$$\left\lceil\max_{1\le i\le n}(b_n+\epsilon)\right\rceil+1.$$
In simpler and more general terms, an open ball is bounded above and so is a finite union of balls. Hence there is a natural larger than the upper bound, which is not covered.
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I don't know, but you in all proof never use diameter so I thnk that this could be use for both metrics – josf Oct 15 '20 at 08:02
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