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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.

Thanks

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    $\begingroup$ 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
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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 $(0,1)$.

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