How could we see that $\{n_k\}_k$ converges $\infty$? Let $x \in \Bbb R\setminus \Bbb Q$ and the sequence $\{\frac {m_k} {n_k}\}_k$  concerges to $x$. The question is from this comment by Ilya:

How could we see that $\{n_k\}_k$ converges $\infty$?

Thanks for your help.
 A: Let $M$ be fixed. Show that there exists such $k_0$ that
$$n_k>M, k\geq k_0.$$
By assuming the contradiction, we'll get such subsequance $\{n_{k_j}\}_j$ that
$n_{k_j} \leq M$ for all $j\geq 1$. Note that
$$\frac{m_{k_j}}{n_{k_j}}\to x.$$ Since such fractions can written as
$$\frac{m_{k_j}}{n_{k_j}}=\frac{A_{k_j}}{M!},$$
where $A_{k_j}$ are integers, then $\frac{m_{k_j}}{n_{k_j}}$ cannot tends to an irrational number. So there is no such subsequance $\{n_{k_j}\}_j$.
A: A sequence of integers if it's convergent then its limit is also an integer and in this case the limit of $(\frac{m_k}{n_k})$ is a rational which contradicts the hypothesis so what we can conclude?
A: If it is bounded, say by $M$, then all the quotients can be written as $A_k/M!$ where $A_k$ is an integer, which clearly does not converge to an irrational.
A: This is actually pretty obvious.
The original conception of an irrational number was an "incommensurable length", that is, one that could not be reconstructed from a segment of unit length. That is, $\frac 5 3$ can be constructed by chopping the unit in 3, and then taking five such subsegments. But no matter how finely you chop up the unit, you will never be able to construct $\sqrt 2$.
If you chop up the unit more and more finely, however, it's obvious that you can get arbitrarily close to any irrational length. However, the key here is more and more finely. If you are unable to chop the unit up into pieces smaller than, say, a hundredth, than obviously there's going to be a limit to your approximation. You'll get yea close using $k$ hundredths, and yea close on the opposite side using $k + 1$ hundredths, and whichever is nearer is as close as you'll get.
Of course, maybe you could get closer using a larger subdivision. But there's only finitely many types of subdivision (in our case, there are 99). So one of them is going to be the one that gets you as close as possible (formally: a finite set always contains a minimal element), and if you want to get closer, you'll have to use finer subdivisions.
It's easy to translate this into a rigorous proof:
Let $\frac 1 n \Bbb Z$ denote the set of rationals with $n$ for their denominator. Obviously there is a minimal $|x - r|$ with $r\in \frac 1 n \Bbb Z$, call this minimum $\delta_n$. $\{\delta_n|n\leq N\}$ is finite and therefore has a minimal element, call it $\epsilon_n$. If the sequence defined in the OP has $n_k$ bounded above by, say, $N$, than in will never be closer to $x$ than $\epsilon_N$, contradicting the definition of convergence.
