$\{ x : x,c \in \mathbb{R} \land c>0 \land \forall j,k \in \mathbb{Z} \ ( k \ge 0 \Rightarrow |x-j2^{-k}| \ge c2^{-k} ) \}$ dense in $\mathbb{R}$?

Question

We define the subset $A\subset \mathbb{R}$ as follows: $x\in A$ if and only if there exists $c>0$ so that $$ |x-j2^{-k}|\geq c2^{-k} $$ holds for all $j\in \mathbb{Z}$ and integers $k\geq 0$. Prove that $A$ is dense

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So I tried showing that for any interval $(a,b)\in \mathbb{R}, (a,b)\cap A \neq \emptyset$. But I am having quite a bit of trouble. Does anyone have a hint on how to proceed?

Answer 1

If $x $ is a rational number wchich can be represent as $\frac{s}{l} $ with $GCD(s,l)=1 $ and such that $2 $ not divide $l$ then $x\in A.$ This is because that $$|2^k x -j|\geq\frac{1}{|l|}$$ for all $j\in \mathbb{Z}.$ But set of such $x$ is dense in $\mathbb{R}.$

Answer 2

$x \in A \Longleftrightarrow$
there exists $c > 0$ with $\forall j \in Z$, integer $n \geq 0$, $c \leq |2^nx - j|.$

$x \notin A \Longleftrightarrow$
$\forall c > 0$, there exists $j \in Z$, integer $n \geq 0$ with $|2^nx - j| < c$ $\Longleftrightarrow$
exists $j \in Z$, integer $n \geq 0$ with $x = j/2^n$
$\Longleftrightarrow x$ is a dyadic rational.

Since $R - Q \subset R - \{ x : x \text{ dyadic rational } \} = A$
and $R - Q$ is dense, $A$ is dense.