Let $x$ be an irrational number. Show that for any positive integer $k ≥ 1$, there exists $δ_k > 0$, such that the open interval $(x − δ_k, x + δ_k)$ does not contain any rational number of the form $\frac pk$ , where p is an integer.
Here is an argument
To prove this statement, we will use the fact that between any two distinct real numbers, there exists a rational number. This fact is known as the density of the rationals in the reals.
Assume that $x$ is an irrational number and let $k$ be a positive integer. We want to show that there exists $δ_k > 0$ such that the open interval $(x - δ_k, x + δ_k)$ does not contain any rational number of the form $\frac pk$, where $p$ is an integer.
Suppose, for the sake of contradiction, that there is no such $δ_k$. Then, for every positive real number $δ$, there exists a rational number of the form $\frac pk$ such that $\left|x - \frac pk\right| < δ$. In particular, there exists a sequence of rational numbers ${p_k}$ such that $p_k = \frac pk$ and $|x - p_k| < \frac 1{k^2}$ for all $k$.
By the density of the rationals in the reals, there exists a rational number $q_k$ such that $x - p_k < q_k < x$. Note that $p_k$ and $q_k$ are both rational numbers of the form $\frac pk$ for some integer $p$, and since $p_k$ and $q_k$ are distinct, we have $q_k - p_k = \frac qk$ for some integer $q$.
Then, we have the following inequalities:
$$x - q_k < p_k - q_k = - \frac qk < 0$$
$$q_k - x < \frac qk$$
Since $x$ is irrational, we know that there exists a positive real number $ε$ such that $|x - y| > ε$ for all rational numbers $y$. We can choose $k$ large enough such that $ \frac 1{k^2} < ε/2$ and $\frac 1k < \frac ε{2q}$. Then, we have:
$$|x - q_k| < \frac qk + \frac 1{k^2} < \frac ε2 + \frac ε2 = ε$$
$$|x - p_k| < \frac 1{k^2} < \frac ε2$$
This implies that $|p_k - q_k| = |(p_k - x) + (x - q_k)| > ε$, which contradicts the fact that $p_k$ and $q_k$ are both rational numbers of the form $\frac pk$ and that $|x - p_k|$ and $|x - q_k|$ are both less than $\dfrac ε2$.
Therefore, our assumption that there is no $δ_k > 0$ such that $(x - δ_k, x + δ_k)$ does not contain any rational number of the form $\dfrac pk$ is false, and the statement is proven.
Please check the solution..or you can give another solution also.