We have a real sequence $\{a_n\}$, in which $a_1=1$ and for $n=p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_r^{\alpha_r}$, $a_n$ defined by $$a_n=\frac{1}{\alpha_1+\cdots+\alpha_r}. $$
I need to find a subset $A$ of $\mathbb{R}$ such that for any $x\in A$, there is a subsequence of $\{a_n\}$ converging to $x$. Thanks for your help.
Define$$ A = \{0\} \cup\left\{ \frac{1}{m} \,\middle|\, m \in \mathbb{N}_+ \right\}. $$
For any $x \in A$, if $x = 0$, take $n_k = 2^k\ (k \geqslant 1)$, then$$ \lim_{k \to \infty} a_{n_k} = \lim_{k \to \infty} \frac{1}{k} = 0. $$ If $x = \dfrac{1}{m}$, take $n_k = p_k^m\ (k \geqslant 1)$, where $p_k$ is the $k$-smallest prime, then$$ \lim_{k \to \infty} a_{n_k} = \lim_{k \to \infty} \frac{1}{m} = \frac{1}{m}. $$
For $x \not\in A$, if $x < 0$ or $x > 1$, then there is obviously no subsequence $\{a_{n_k}\}$ such that $a_{n_k} \to x$ $(k \to \infty)$.
If $0 < x < 1$, since $x \not\in A$, suppose $\dfrac{1}{m + 1} < x < \dfrac{1}{m}$, where $m \in \mathbb{N}_+$. Because for any $n \in \mathbb{N}_+$, there is $a_n \in A\setminus\{0\}$, then$$ |a_n - x| \geqslant \min\left( \left| \frac{1}{m} - x \right|, \left| \frac{1}{m + 1} - x \right| \right) > 0. $$ Thus there is no subsequence $\{a_{n_k}\}$ such that $a_{n_k} \to x$ $(k \to \infty)$.