# Sequence in $(0,1)$ without accumulation points

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

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