Proving that a set has empty interior For $n=0,1,2...$ define $$A_n = \bigcup^{3^n}_{k=1}\left(\frac{k-\frac23}{3^n},\frac{k-\frac13}{3^n}\right)$$ Let $A=\bigcup_{n=0}^{\infty}A_n$, and let $B=[0,1]\cap \overline A$. I wish to show that $B$ has an empty interior. First, I suppose it has an interior. So I let $x\in B$ and $0\leq x \leq \frac{k-\frac23}{3^n}$ and assume the existence of $\delta>0$ such that $(x-\delta,x+\delta)\subset B$. Then I tried to choose $n$ such that $\frac{\frac 16}{3^n}<\delta$ then add $\frac{\frac 16}{3^n}$ to $x$ to yield $\frac{\frac 16}{3^n}<x+\frac{\frac 16}{3^n}<\frac{k-\frac12}{3^n}$ which is in $A_n$ for k=1. Is this the right direction of the proof? Thank you in advance.
 A: The direction of the proof is great but it needs some tiny adjustments. In short, we should choose $n$ after $\delta$.
Suppose $x\in B$ and assume the existence of $\delta>0$ such that $(x-\delta,x+\delta)\subset B$. Then we can choose $n$ such that $\frac{\frac 13}{3^n}<\delta$. Since $x\in B$, there exists $k$ such that $\frac{k-\frac 13}{3^n}\leq x \leq \frac{k +1-\frac 23}{3^n}=\frac{k +\frac 13}{3^n}$.
Now there are two cases.
If $\frac{k-\frac 13}{3^n}\leq x \leq \frac{k}{3^n}$, then we minus $\frac{\frac 13}{3^n}$ from $x$ and get $\frac{k-\frac 23}{3^n}<x-\frac{\frac 13}{3^n}<\frac{k-\frac 13}{3^n}$. Thus $x-\frac{\frac 13}{3^n}\in A_n$.
If $\frac{k}{3^n}\leq x \leq \frac{k + \frac 13}{3^n}$, then we add $\frac{\frac 13}{3^n}$ to $x$ and get $\frac{k+\frac 13}{3^n}<x+\frac{\frac 13}{3^n}<\frac{k+\frac 23}{3^n}$. Thus $x+\frac{\frac 13}{3^n}\in A_n$ (for $k+1$).
But we already assumed that $(x-\delta,x+\delta)\subset B$ and $\frac{\frac 13}{3^n}<\delta$. Such contradiction proves the claim.
