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Let $X$ be the integers with metric $ρ(m,n)=1$, except that $ρ(n,n)=0$. Check that $ρ$ is a metric. Show that $X$ is closed and bounded, but not compact.

This is a "made-up" example demonstrating closed and bounded doesn't imply compactess in more general metric space. I checked that $\rho$ is a metric already. Yet I have no idea how to approach "showing $X$ is closed and bounded." I visualized this metric to be a set of number with just $0$ and $1$ (or maybe this is not correct?). Also, I doubt that this metric is NOT compact. Anyway, I'd appreciate if you can help! Meanwhile, do we use a ball $B(0,1)$ in general to show that a metric is closed and bounded? If so, why?

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  • $\begingroup$ Get the defiinitions from a note or wikipedia, that can help.. $\endgroup$ – Berci Mar 29 '13 at 1:50
  • $\begingroup$ I find that the counter example is clear, but I fail to understand the conceptual idea behind it. What example is wrong with metric spaces that fail to be compact if bounded and closed? Why is this suppose to be obvious? $\endgroup$ – Charlie Parker Mar 6 '18 at 15:19
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Your metric generates the discrete space where every subset of $X$ is open (and thus also closed). It is bounded, because each point lies within a distance $1$ of some point $x_0$ (any will do). It is not compact, because $\{\{x\} \mid x \in X\}$ is an open cover of $X$, but you won't be able to pick finite subcover, because $X$ is infinite.

I hope this helps ;-)

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    $\begingroup$ I am Astonished! Many many thanks!! I stayed in a continuous space and wondered for about an hour! It's like an epiphany! $\endgroup$ – user48601 Mar 29 '13 at 1:56
  • $\begingroup$ I find that the counter example is clear, but I fail to understand the conceptual idea behind it. What example is wrong with metric spaces that fail to be compact if bounded and closed? Why is this suppose to be obvious? $\endgroup$ – Charlie Parker Mar 6 '18 at 15:19
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    $\begingroup$ @CharlieParker I have no idea what are you referring to. There is nothing wrong with such spaces or examples of such spaces (and there is nothing "right" about them either, they just are/happen and that's it). What is supposed to be obvious (you were the first to use that word here)? $\endgroup$ – dtldarek Mar 6 '18 at 15:48
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Here is a simple example. Denote by $\ell^\infty$ the set of all bounded sequences of real numbers; put $$d(x,y) = \sup_n |x_n - y_n|.$$ Then all sequences of distance $\le 1$ from the zero sequence is closed, bounded but it is not compact.

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  • $\begingroup$ Oh, thanks for your quick reply. I'd centainly agree that this is an example. But I am confused the one I mentioned in my original post. It's a example in my textbook but I don't understand this example in particular. Thanks! $\endgroup$ – user48601 Mar 29 '13 at 1:52
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    $\begingroup$ I think this example might look less artificial to you than that of the discrete metric. $\endgroup$ – ncmathsadist Mar 29 '13 at 1:56
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By Riesz's lemma, we know that the unit ball which is closed and bounded of an infinite-dimensional normed space (which is a particular case of metric space) is never compact.

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