Prove there is no sequence of real numbers for which the set of subsequential limits is preciesely $(0,1)$.
I have seen the similar question that shows there exists a sequence of real numbers for which the subsequential limits is $[0,1]$ like the following:
Existence of sequence whose set of subsequential limits is $[0,1].$
In the similar problem above, we can give a concrete example, but here we need to disprove the existence of such sequence. How do I do that?
Suppose such a sequence $(a_k)_k$ of real numbers exists. By assumption, $1/n$ is a sub-sequential limit, so there exists some $k_n$ such that $|a_{k_n} - \frac{1}{2n}| < \frac{1}{2n}$. Then $|a_{k_n}| < \frac{1}{n}$.
By choosing $k_n$ sequentially for each $n=1,2,\ldots$ such that $k_1 < k_2 < \cdots$, we will have produced a subsequence $(a_{k_n})_n$ such that $|a_{k_n}| < 1/n$, so $a_{k_n} \to 0$.
If you can get arbitrarily close to $1/n$ for every $n$ then you can get arbitrarily close to $0$, so no such sequence can exist.
