We know that the one-point compactification of $\mathbb{Z}_{+}$ is homeomorphic with the subspace $\{ 0 \} \cup \{ 1/n \mid n \in \mathbb{Z}_{+} \}$ of $\mathbb{R}$.
I have for all $1≤k<+∞$ the following inequality:
$$p_{k}≤1/x_{k}.............(*)$$ where $p_{k}$ is the prime sequence and $0<x_{k}<1$ is a strictly positive real sequence without lower bound, i.e., there is no real number $a>0$ verifying $a<x_{k}$ for all $1≤k<+∞$. Here I understand that the inequality holds true for all finite primes. If $k=+∞$, then $x_{k}=0$.
But since $0<x_{k}<1$ is bounded, then in the usual topology of $\mathbb{Z}_{+}$ inequality $(*)$ cannot holds for infinitely many primes.
My question is: By this one-point compactification of $\mathbb{Z}_{+}$. Is it possible to say that every prime number is located in the interval $[2,1/x_{k}]$