An Inequality related to the Cantor Set $\mathscr C \subset [0,1]$ 
Let $\mathscr C\subset\mathbb R$ be the Cantor set on the interval $[0,1]$. Let $x\in \mathscr C$, and $0 < r < 1$ such that $$\frac{2}{3^k} < r \le \frac{2}{3^{k-1}}$$
for some positive integer $k$. Recall that $\mathscr C = \bigcap_{k=1}^\infty E_k$, where $E_k$ consists of intervals of length $3^{-k}$.

*

*Show that the interval $B(x,r) := [x-r,x+r]$ overlaps one of the intervals in the complement of $E_{k-1}$ by at least $$\delta = \min\{r - 3^{-k}, 3^{-(k-1)}\}$$

*Further, show that $\delta \ge \frac r 2$.


My work.
Assuming $(1)$ holds, I could show $(2)$.

*

*By the choice of $k$, $3^{-(k-1)}\ge \frac r 2$.

*Also, $r > 2\cdot 3^{-k}$ implies that $\frac r 2 > 3^{-k}$. Subtracting $r$ on both sides, $r - 3^{-k} \ge \frac r 2$.

Therefore, $\delta \ge \frac r2$. Could someone help me show $(1)$?
A pictorial explanation or something along those lines may help. Thanks a lot!
 A: I prefer to give an explaination with some figures. In the following two lines, I try to show you what $E_k$ and $E_{k-1}$ looks like on the same part of the real axis. "[--$E_k$--]" represents a closed interval in $E_k$, while "(--$\mathbb{R}\setminus E_k$--)" represents an open interval of $\mathbb{R}\setminus E_k$. Note that the long open interval on the right is contained in both $\mathbb{R}\setminus E_k$ and $\mathbb{R}\setminus E_{k-1}$.

[--$E_k$--](-$\mathbb{R}\setminus E_k$-)[--$E_k$--](------$\mathbb{R}\setminus E_k$------)
[-------------$E_{k-1}$-----------](-----$\mathbb{R}\setminus E_{k-1}$-----)
Since $x\in \mathscr{C}$, we have $x\in E_k$. Suppose $x$ lies in the right interval of $E_k$ in the figure. Then the length of $[x-r,x+r]\setminus E_{k-1}$ is at least $r-3^{-k}$ (It reaches the minimum $r-3^{-k}$ when $x$ is the left endpoint.) Since the intervals of $\mathbb{R}\setminus E_{k-1}$ has a minimum length $3^{-(k-1)}$, the overlapping length is at least $\min\{r-3^{-k},3^{-(k-1)}\}$.
