Variation on the Cantor Set

We're going to create a set $\displaystyle C$ that is a variation on the Cantor Set. Start with the open interval $\displaystyle C_0=(0,1)$. Remove the segment $\displaystyle \left[\frac{1}{3},\frac{2}{3}\right]$. Now you have $\displaystyle C_1=\left(0,\frac{1}{3}\right)\cup\left(\frac{2}{3 },1\right)$. Next remove $\displaystyle \left[\frac{1}{9},\frac{2}{9}\right]$ and $\displaystyle \left[\frac{7}{9},\frac{8}{9}\right]$ and you're left with $\displaystyle C_2=\left(0,\frac{1}{9}\right)\cup\left(\frac{2}{9 },\frac{1}{3}\right)\cup\left(\frac{2}{3},\frac{7} {9}\right)\cup\left(\frac{8}{9},1\right)$. Repeat ad infinitum...

I suspect this set is empty. Proof:

Want to Prove: There are no interior points.

Assume that $\displaystyle \exists~ x\in C_n, \exists~ \epsilon>0$ such that $\displaystyle N(x,\epsilon)\subset C_n$.

For $\displaystyle C_0, \epsilon<\frac{1}{2}$ to create a neighborhood contained in $\displaystyle C_0$.

For $\displaystyle C_1, \epsilon<\frac{1}{2}\cdot\frac{1}{3}$.

For $\displaystyle C_2, \epsilon<\frac{1}{2}\cdot\frac{1}{9}$.

For $\displaystyle C_n, \epsilon<\frac{1}{2\cdot3^n}$.

Thus, as $\displaystyle n\to\infty$, we have that $\displaystyle \epsilon<0$, which is clearly a contradiction. Thus there are no interior points in this set. However, $\displaystyle C$ is an open set (I'm worried this is false), so all points are interior points, leading us to the conclusion that $\displaystyle C$ is empty.

Is this correct?