5
$\begingroup$

In Rudin, it says the Heine-Borel theorem holds for Euclidean metric spaces.

What is an example of a metric space where Heine-Borel does not hold true?

$\endgroup$
  • $\begingroup$ From en.wikipedia.org/wiki/Heine%E2%80%93Borel_theorem: the metric space of rational numbers (or indeed any incomplete metric space) fails to have the Heine–Borel property $\endgroup$ – stewbasic Oct 4 '17 at 23:11
  • 1
    $\begingroup$ Any metrizable topological space has a bounded metric. $\endgroup$ – bof Oct 4 '17 at 23:16
  • 1
    $\begingroup$ The point of the commrnt from bof is that if $(X,d)$ is a non-compact metric space (E.g. $X=\Bbb R$ with the usual metric) and if $e(x,y)=\min (1,d(x,y)) $ then the metric $e$ generates the same topology that $d$ does. So with respect to the metric $e,$ the set $X$ is bounded, and it is closed, but not compact. $\endgroup$ – DanielWainfleet Oct 5 '17 at 1:34
8
$\begingroup$

Consider $\Bbb R^2 \setminus \{(0,0)\}$ with the usual metric restricted from $\Bbb R^2$. The set $$D = \{ (x,y) \in \Bbb R^2 \mid 0 < x^2+y^2 \leq 1 \}$$is closed in $\Bbb R^2 \setminus \{(0,0)\}$, bounded, but not compact. Sequences in $D$ which "want" to converge to $(0,0)$ don't have limit in $D$.

$\endgroup$
4
$\begingroup$

Any non-complete metric space, or an infinite-dimensional Banach space.

$\endgroup$
  • 1
    $\begingroup$ The non-complete examples are almost cheating because you are inclined to say "well sure but then I complete it and the problem goes away, right?". (Or alternately you are inclined to replace the word "closed" in Heine-Borel with "complete", seeing as the two are the same for subspaces of a complete metric space.) The infinite dimensional Banach space examples are very important by comparison. +1 for that. $\endgroup$ – Ian Oct 4 '17 at 23:21
3
$\begingroup$

Motivated by a comment on another answer, here goes another example (which avoids the psychologically problematical "non-complete" case, while being elementary):

Let $(M,d)$ be an infinite set with the discrete metric. $M$ is then bounded, obviously closed as a subset of itself, but is not compact (the cover $\{x\}_{x \in M}$ is an open cover with no finite subcover). Note that $M$ is also complete.

$\endgroup$
1
$\begingroup$

Consider the space $$ c_0=\{(x_n)_n:\ \lim_nx_n=0\}, $$ with the distance $$d(x,y)=\sup\{|x_n-y_n|:\ n\in\mathbb N\}.$$

The unit ball $$B=\{x\in c_0:\ d(x,0)\leq1\}$$ is closed, bounded, and it contains the sequence $$E=\{e^k:\ k\in\mathbb N\},$$ where $e^k$ is the sequence with entries $\delta_{k,n}$, i.e., $e^k$ consists of all zeroes and a one in the $k^{\rm th}$ position. It is easy to see that $$d(e^k,e^h)=1,\ \ h\ne k,$$ so the sequence $E$ admits no convergent subsequence.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.