In $\mathbb R$, a subset is compact in and only if it is closed and bounded. The open interval $A=(0,1)$ is not closed so is not compact. Hence there must exist a sequence $\{a_n\}_{n\in\mathbb N}$ without cluster points (otherwise $A$ would be compact). Is it possible to find such a sequence?

It should be easy since the sequence has countable many points where the open interval is uncountable, but I can't find any sequence. On the other hand, after having a look at If $X$ is not countably compact, then there exists a countable subset without accumulation points, I'm not sure this is possible, because $A$ is countable compact, isn't it?

Motivation. I'm studing sequential spaces and I was wondering how the definition of sequentially open is modifed when you allow the sequence to have no limit points. I have included the tag soft questionbecause I'm not really worried about that. It's out of curiosity.


  • 1
    $\begingroup$ Any sequence in $(0,1)$ which is convergent to $0$ or $1$ elements of $\bar{A}\setminus A$ (cluster points outside $A$). $\endgroup$ – Robert Z Oct 3 '18 at 16:57
  • $\begingroup$ Mmmm yes. It was obvious, wasn't it? Anyway, thanks. $\endgroup$ – Dog_69 Oct 3 '18 at 17:06
  • $\begingroup$ What do you mean? Did I misunderstood your question? $\endgroup$ – Robert Z Oct 3 '18 at 17:09
  • $\begingroup$ @RobertZ No no. But thanks to your comments I have realized that it was an stupid question. But your answer is fine. $\endgroup$ – Dog_69 Oct 3 '18 at 17:15

The sequence defined by $a_n = \frac{1}{n}$ for $n \geq 2$ is a sequence of $(0,1)$ that has no accumulation point in $(0,1)$.


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.