# Fractal Dimension of $C_{\frac{1}{3}}\times[0,1]$

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 $$\dim A+\dim B<\dim A\times B < \dim A + \dim B +1$$

• It can hardly be anythink else then $1+\ln2/\ln 3$.. – Peter Franek Dec 14 '14 at 18:05
• why is that correct? – user65985 Dec 14 '14 at 18:08
• I appologize for being lazy to come up with a formal proof, but in all reasonable cases, all reasonable definitions of dimensions, the dimension increases by one if you multiply a set with $[0,1]$. – Peter Franek Dec 14 '14 at 18:09
• Much wilder sets are needed for the strict inequality you want. From wikipedia: "...it is known that when $X$ and $Y$ are Borel subsets of $R^n$, the Hausdorff dimension of $X \times Y$ is bounded from above by the Hausdorff dimension of $X$ plus the upper packing dimension of $Y$." – Peter Franek Dec 14 '14 at 18:17

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.
• I am reading a book by Furstenberg, he gave two examples.Another more tricky one is constructed on $[0,1]$: let $A=\{\sum_{i\in N_1}\frac{a_i}{2^i}:a_i\in\{0,1\}\}$,let$B=\{\sum_{i\in N_2}\frac{a_i}{2^i}:a_i\in\{0,1\}\}$, where $N_1=\cup_{\mbox{n is even}}[n!,(n+1)!)\cap \mathbb N$,$N_2$ is $\mathbb{ N}- N_1$. Then $dim(A\times B)\geq 1$, but $dim(A)=dim(B)=0$. – user201043 Dec 15 '14 at 1:34
• What is the name of the book? I also thought about a similiar example but couldn't find out if and why $\dim(A\times B)\ge 1$. Also I want $A,B$ which satisfy both the LHS inequality and the tricot inequality from the RHS. – user65985 Dec 15 '14 at 4:20