I wonder what is the dimension of the fractal set given by the product of the unit interval $[0,1]$ by the thirds-cantor-set ($C_\frac{1}{3}=\bigcap_n C_n$ where $C_0=[0,1],C_1=[0,\frac 1 3]\cup[\frac 2 3, 1]$ and so on).
On the one hand it seems like it's dimension is 2 since every $C_n\times[0,1]$ is a union of (disjoint) rectangles which have non empty interior but I'm not sure since the interior of the cantor set in $\mathbb{R}$ is empty.
How can I find the exact Hausdorff dimension of $C_\frac 1 3\times[0,1]$ ?
EDIT: more generally I'm looking for two sets $A,B$ s.t $$\dim A+\dim B<\dim A\times B < \dim A + \dim B +1$$
We already know the Hausdorff dimension of $C_{\frac{1}{3}}$ is $\frac{log2}{log{3}}$, furthermore I find something from here, we can apply the result to conclude that $dim_{H} (C_{\frac{1}{3}}\times[0,1])=1+\frac{log2}{log3}$. For the example such that $dim_HA+dim_{H}B<dim_H(A\times B)$, we may consider the standard one dimensional Brownian motion, think its paths on the plane and its projection to$[0,1]$, since we have the graphs of its paths has dimension $\frac{3}{2}$ almost surely.