Is this set meager? $A = \{x\in \mathbb{R}: \exists c>0, |x-j2^{-k}|\geq c2^{-k}, \forall j\in \mathbb{Z}, k\geq 0 \}$

Question

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

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I am completely lost on this one. It looks like it is saying that $x\in A$ if the difference between $x$ and any dyadic rational can be made greater than $c2^{-k}$. But I have no intuition for this at all.

Anyone have a hint (not a solution) for how to approach this problem?

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Edit:

So I tried to see if for each $k$, say $0,1,2,...$, I could create a set $A_k$ which is nowhere dense. But $A_0 = \mathbb{R}/\mathbb{Z}$ is not nowhere dense. So that approach isn't working unless I am confused.

Answer 1

We have $A=\bigcup A_n$, where $$A_n=\{x\in \mathbb{R}: |x-j2^{-k}|\geq 2^{-n-k}, \forall j\in \mathbb{Z}, k\geq 0 \}.$$

Spoiler:

So it suffices to show that each $A_n$ is meager. Since $$A_n=\bigcap_{j\in\Bbb Z, >! k\ge 0} \Bbb R\setminus (j2^{-k}-2^{-n-k}, j2^{-k}+2^{-n-k}),$$ it is closed. Since $A_n$ is disjoint with the set of diadic rationals, the set $A_n$ is nowhere dense.