We know that every closed and bounded subset of $\Bbb{R}$ is compact. The proof proceeds by bifurcating $[a,b]$, and then using the property that in a complete metric space the infinite intersection of closed and bounded sets contains one point.

I was wondering if this can be extended to any complete metric space. Is a closed and bounded set of any complete metric space compact? If one were to extrapolate the proof given above for this situation, how do you bifurcate a complete metric space?

Also, assume that the complete metric space is ordered. Can you still bifurcate it? How do you find the mid-point?

Thanks in advance!

  • 2
    $\begingroup$ No. See here. Additionally, what does it mean to be an "ordered metric space"? Is the order in some way compatible with the metric? If so, how? If not, then (by the Order-Extension Principle), we can always order a metric space, and it makes no difference. $\endgroup$ – Cameron Buie Aug 29 '13 at 4:07

Let $X$ be an infinite set. Define a metric $d$ on $X$ by $d(x,y) = 0$ if $x = y$, and $d(x,y) = 1$ if $x \neq y$. This is known as the discrete metric. This metric makes $X$ a complete metric space.

Let $E$ be any infinite subset of $X$. Then $E$ is closed and bounded, but not compact.

Remark: If we assume in addition that the set is totally bounded, the usual proof can be carried out. Therefore closed and bounded subset of a complete metric space need not be compact.

  • $\begingroup$ So. Much. Simpler. Big ol' +1 from me. $\endgroup$ – Cameron Buie Aug 29 '13 at 4:10

André’s example is probably the simplest; here’s a fairly general technique for constructing such examples. If $\langle X,d\rangle$ is any metric space, the function

$$d\,':X\times X\to\Bbb R:\langle x,y\rangle\mapsto\min\{d(x,y),1\}$$

is a metric on $X$ that generates the same topology as $d$. If $\langle X,d\rangle$ is complete but not compact, $\langle X,d\,'\rangle$ provides a counterexample: every subset of $X$ is $d\,'$-bounded, so $X$ is a closed, bounded subset that is not compact. Since $\Bbb R$ with the usual metric is complete but not compact, it provides one such example.


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