Here is another question I have concerning prime numbers. Let $p_n$ denote the $n$'th prime number. For all $\alpha \in (0,1)$ it can be proven that exists infinitely many $k \in \mathbb{N}$ such that $$p_{n_k+1} - p_{n_k} < p_{n_k}^{1-\alpha}$$ Let $\alpha_n \to 1$ increasing, and define: $$ C_{n,1} = \left\{ p_{n_k}| p_{n_k+1} - p_{n_k} < p_{n_k}^{1 - \alpha_n}\right\} $$ Obviously $C_{n+1,1} \subseteq C_{n,1}$ for all $n \in \mathbb{N}$. Now I think $C_{n,1}$ is a compact subset of $\mathbb{R}$ for all $n$ hence using Cantor's intersection theorem we have $$ \cap_{n \in \mathbb{N}} C_{n,1} \neq \emptyset$$ Let $q_{n_1}$ be the smallest element form this intersection. Now of course, define
$$ C_{n,2} = \left \{ p_{n_k} > q_{n_1} | p_{n_k+1} - p_{n_k} < p_{n_k}^{1 - \alpha_n} \right\}$$ Again $$ \cap_{n \in \mathbb{N}} C_{n,2} \neq \emptyset$$ hence let $q_{n_2}$ be the smallest, etc. In such a way $(q_{n_k})_{k\in \mathbb{N}}$ is formed with the property that $$ q_{n_k+1} - q_{n_k} < q_{n_k}^{1-\alpha_n}$$ $\forall n \in \mathbb{N}$.
Is everything correct?
Edit:
Those subsets are not compact (they are indeed closed, but are not bounded) hence there is a big error!
