Let $A=\{a_k:k \in \mathbb{N}\}$ be a countable set of strictly positive real numbers such that $\mbox{inf}(A) = 0$. Prove that $0$ is an accumulation point of A. Must $0$ be the only accumulation point of $A$?
A point is an accumulation point if there exists an $x_n \in A$ with $x_n \ne x$ for any $n$ such that $x_n \rightarrow x$. Intuitively, it is clear that $0$ is an accumulation point since the infimum is zero yet the set is strictly positive. I'm not sure how to rigorously prove this. If I try to show that the limit of $a_k$ is zero then I think that would rule out whether there can be another accumulation point, and I feel that there is another one.
For $k\in\Bbb N$, let $n_k$ be the smallest integer with $a_{n_k}<\frac1k$ (which exists by the given infimum). Then $a_{n_k}\to 0$, showing that $0$ is an accumulation point.
On the other hand, $A$ might be the countabke set $\Bbb Q\cap(0,\infty)$, which has every $a\ge0$ as accumulation point.