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
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 $(0,1)$.