How to show that $\bigcup_{n \in \mathbb{N}} \{ \frac{k}{2^n} : 1 \leq k \leq 2^n \}$ is dense in $\mathbb{Q} \cap [0,1]$?

Question

I want to show that some subsets of the rationals are dense in $[0,1]$. Right now I have the set $$P =\bigcup_{n \in \mathbb{N}} \{ \frac{k}{2^n} : 1 \leq k \leq 2^n \}$$ If I can show that this is dense in $\mathbb{Q} \cap [0,1]$ then I'm done since $\mathbb{Q} \cap [0,1]$ is dense in $[0,1]$. I want to find a sequence $(x_n)_n \subset P$ which converges to an element $\frac{p}{q} \in \mathbb{Q} \cap [0,1]$.

Answer 1

This set $P$ is called a set of dyadic numbers. Let $x\in[0,1]$. Put $\varepsilon>0$. Since $\frac{1}{2^n}\to 0$, there exists $n\in\Bbb N$ s.t. $\frac{1}{2^n}<\varepsilon$. Moreover, imagine the partition of $[0,1]$ into $2^n$ equal parts, each of diameter less than $\varepsilon$. Our $x$ is in one of subintervals of this partition. That is all.

I have shown that $P$ is dense in the whole interval $[0,1]$, not only in rationals of $[0,1]$.

My method shows that arbitrarily close to $x$ there is a dyadic number. The method for rationals is the same.