# Example of a metric space where Heine-Borel theorem does not hold

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?

• 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 – stewbasic Oct 4 '17 at 23:11
• Any metrizable topological space has a bounded metric. – bof Oct 4 '17 at 23:16
• 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. – DanielWainfleet Oct 5 '17 at 1:34

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 $D$.

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

• 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. – Ian Oct 4 '17 at 23:21

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.

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.