Given $x \in \mathbb{R}$ and $n \in \mathbb{N}$ there exists $k \in \mathbb{Z}$ such that $\left|2^{n} x-k\right|<1$ (this has been proven on a previous question).
Deduce from this that $\left\{\frac{k}{2^{n}}: k \in \mathbb{Z} \quad n \in N\right\}$ is dense in $\mathbb{R}$
I have tried using the Archimedean property of Real numbers but I do not know where to go from there. I am aware that my final objective is to show that $x, y \in \mathbb{R} \quad x<y$ s.t.$\quad x<\frac{k}{2^{n}}<y$
For the given two numbers $x<y$, let $\epsilon=y-x$.
According to Archimedean property, there exist natural number k such that $\frac{\epsilon}2\cdot k>1$. $2^k>k$ is so trivial.
Then, $2^ky-2^kx>2$, so there exists an natural number $n$ such that $2^kx<n<2^ky$.
So $x<\frac n{2^k}<y$. The proof ends.
HINT: Then given $x \in \mathbb{R}$ and $n \in \mathbb{N}$, there exists a $k \in \mathbb{Z}$ such that $|x-\frac{k}{2^n}| \le 2^{-n}$. So for any $x \in \mathbb{R}$, you can make $n$ arbitrarily large.
So now let $x_1$ and $x_2$ be 2 elements in $\mathbb{R}$ satisfying $x_1 \not = x_2$. It suffices to show that there is an element in the set $\{\frac{k}{2^n}; k \in \mathbb{Z}; n \in \mathbb{N}\}$ between $x_1$ and $x_2$. Now let $x=\frac{x_1+x_2}{2}$. By the above paragraph, for every $n \in \mathbb{N}$, there exists a $k \in \mathbb{Z}$ such that $|x-\frac{k}{2^n}| \le 2^{-n}$. What if you were to use the above paragraph, and make $n$ large enough so that $2^{-n} < \frac{|x_2-x_1|}{4}$? Then you note that there is a $k$ such that $|x-\frac{k}{2^n}| \le \frac{|x_2-x_1|}{4}$, which gives $\frac{k}{2^n}$ must fall between $x_1$ and $x_2$...
You mentioned that it suffices to prove the following proposition I'll call (P): $$\text{(P)}\qquad\text{if $x<y$, there exist $k,n$ such that $x<\frac{k}{2^n}<y$.}$$ If (P) holds, then certainly the dyadic rationals are dense in $\mathbb R$. Here is another property (Q) that implies the dyadic rationals are dense, and follows more immediately from what you have already shown: $$\text{(Q)}\qquad\text{for each $\varepsilon>0$, and each $x\in\mathbb R$, there exist $k,n$ such that $|x-\frac k{2^n}|<\varepsilon$.}$$ Property (Q) says that any interval centered at $x$ contains a dyadic rational, which is a way of stating density of the collection of dyadic rationals that generalizes more easily to other dimensions/spaces. You can prove (Q) immediately from what you already know by using the archimedean property of $\mathbb R$ to find $n$ such that $2^n>1/\varepsilon$.
You can also prove (Q) implies (P) (in fact, they are equivalent). To do this, you can apply (Q) to the midpoint of $x$ and $y$ and an appropriate choice of $\varepsilon$.
An alternative without using the given statement:
Note that $0$ is a limit point of the given set (because the sequence $(x_n)$ where $x_n=\frac 1{2^n}\in$ the given set, converges to $0$).
Let $a<b$ be arbitrarily chosen. Let $\epsilon =b-a$. For large enough $n$, we have $\frac 1{2^n}\lt \epsilon \tag 1$
Define $M:=\{m\in \mathbb Z: \frac m{2^n}\le a\}$. Note that $M$ is non-empty and is bounded above by $\epsilon$ so must have a supremum $p\in \mathbb Z$. Hence, $\frac {p}{2^n}\le a$ and $\frac {p+1}{2^n}\gt a$
Claim: $\frac {p+1}{2^n}\lt b$.
Proof: Suppose on the contrary that, $\frac {p+1}{2^n}\ge b$, it follows that $\frac {p+1}{2^n}-\frac {p}{2^n}\ge b-a\implies \frac 1{2^n}\ge \epsilon$, which contradicts $(1)$. So $\frac{p+1}{2^n}\in (a,b)$. This completes the proof.
