Dimension of Cantor set with middle quarter removed

Question

So in this Cantor set the middle fourth is removed at each stage. I have shown that the length of this set is 0. I want to verify that the dimension I have calculated is the right one. $$C_1=\left[0, \frac38\right] \cup \left[\frac58, 1\right]$$ In order to get two Cantor set we should multiply the Cantor set by $\frac83$ so we have dimension $d$ $$d=\frac{\ln2}{\ln(8/3)}=0.7066$$

Answer 1

If you are talking about the Hausdorff dimension, the standard way of computing it for a cantor-like set is to exploit self-similarity and to rewrite it as union of two properly scaled and traslated version of itself. As an example, in your case $$C_1=\frac{3}{8}C_1\cup(\frac{3}{8}C_1+\frac{5}{8}) .$$ From which, if $0<H^d(C)<\infty$, $$1=2\left(\frac{3}{8}\right)^d\implies d=\log_{\frac{8}{3}}2,$$ which is your result. Anyway the subtle (and not so easy) point is to show that $H^d(C)$ is neither $0$ nor $\infty$.