Prove that a countably infinite set of positive real number with finite non zero limit point can be arranged in a sequence $(a_n)$ s.t $(a_n^{1/n})_n$ is convergent.
I am not getting the proof:
Let $(x_n)$ denote the real number and let $a$ be a finite non zero limit point. Choose $A$ s.t $1/A < a< A$ and let $a_1$ denote the $x_n$ of smallest subscript that lies in the interval $(1/A,A)$. Now I think that I have to go by induction that chosen $a_1,...,a_{k-1}$, I have to choose $a_n$ but I am not able to get the result.
Let $S=(x_n)_{n\in \Bbb N}$ be a positive real sequence with $x$ being a positive limit point of $S.$ Let $f:\Bbb N\to \Bbb N$ be strictly increasing such that $x=\lim_{n\to \infty}x_{f(n)}$ and such that $\Bbb N \setminus f(\Bbb N)$ is infinite. This is possible because $S$ has a sub-sequence $(x_{n_j})_{j\in \Bbb N}$ converging to $x$, and we may let $f(\Bbb N)=\{n_{(2j)}:j\in \Bbb N\}.$
Let $g:\Bbb N\to \Bbb N \setminus f(\Bbb N)$ be the (unique) order-isomorphism. Take a strictly increasing $h:\Bbb N \to \Bbb N$ such that $h(n+1)\geq h(n)+2$ and where $h(n)$ is large enough that $|(x_{g(n)})^{1/h(n)}-1|\le1/n.$
Let $i:f(\Bbb N)\to \Bbb N\setminus h(\Bbb N)$ be the (unique) order-isomophism. Finally if $m=i(f(n))$ let $y_m=x_{f(n)}$ and if $m=h(n)$ let $y_m=x_{g(n)}.$