I was reading through the Rosenlicht Analysis Text and I could not think of a subset $S$ of metric space $E$ where $S$ is bounded and closed but not compact. Could someone give me an example of this and why it's true? Thanks!

I was thinking that the subset $[0,1]$ of the metric space $R$ is closed and bounded however it is not compact.

  • $\begingroup$ maybe the set of two distinct points? $\endgroup$ – Emilio Novati Feb 17 '15 at 19:34
  • $\begingroup$ @EmilioNovati so like [0,1] right? $\endgroup$ – SilverCat Feb 17 '15 at 19:34
  • $\begingroup$ No! this is not a set with only two distict points $\endgroup$ – Emilio Novati Feb 17 '15 at 19:35
  • $\begingroup$ Isn't the set with the points from 0 to 1, [0,1] closed and bounded? How is [0,1] compact? $\endgroup$ – SilverCat Feb 17 '15 at 19:38
  • $\begingroup$ @SilverCat It depends on the metric. We have a famous theorem (Heine Borel) that tells us in $\Bbb R^{n}$ with the Euclidean metric, a set is compact if and only if it is both closed and bounded. $[0,1]$ is closed and bounded, so by the theorem it must be compact. $\endgroup$ – layman Feb 17 '15 at 19:40

Equip $\Bbb R$ with the discrete metric, i.e., $d(x,y) = \begin{cases} 0 & x = y \\ 1 & x \neq y \end{cases}$.

Then the interval $[0,1]$ is a subset of $\Bbb R$ which is closed (since its complement is open), bounded since $[0,1] \subseteq B(\frac{1}{2}, 2)$ (where $B(x,\epsilon)$ is the ball around $x$ of radius $\epsilon$), but the set is not compact, since the collection of singletons $\{x \}$ for each $x \in [0,1]$ is an open cover with no finite subcover.

  • $\begingroup$ Please don't accept this answer just yet. Wait a short while for others to answer because someone might have some valuable input, or may produce a better answer. $\endgroup$ – layman Feb 17 '15 at 19:41
  • $\begingroup$ Awesome so this would be the subset that is closed and bounded but not compact. Could we define a metric space that is closed and bounded and not compact that has the above subspace that is closed and bounded and not compact? $\endgroup$ – SilverCat Feb 17 '15 at 19:42
  • $\begingroup$ @SilverCat I mean, you could take the interval $[-5, 5]$ with the discrete metric. It has as a subspace the example above, and neither the subspace nor it are compact, but both are closed and bounded. $\endgroup$ – layman Feb 17 '15 at 19:43
  • $\begingroup$ Thank you! Makes perfect sense! $\endgroup$ – SilverCat Feb 17 '15 at 19:46

$[0,1]\cap \mathbf Q$ is a bounded, closed subspace of the metric space $\mathbf Q$. However, it is not not compact since it is not complete.

  • $\begingroup$ You mean it's closed as a subspace of $\Bbb Q$. $\endgroup$ – layman Feb 17 '15 at 20:02
  • $\begingroup$ Well, it's a metric space. If you're in $\mathbf R^n$, a closed bounded subspace is compact. $\endgroup$ – Bernard Feb 17 '15 at 20:07
  • $\begingroup$ I'm just filling in details that the OP may not know how to fill in, not questioning the validity of your answer. $\endgroup$ – layman Feb 17 '15 at 20:09
  • $\begingroup$ @user46944: Oh! Sorry I misunderstood your comment. You're right, I'll add that. $\endgroup$ – Bernard Feb 17 '15 at 20:11

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.