Hausdorff dimension of Cantor middle third set Can anyone please explain me the part (see picture) of calculation of lower estimate for Cantor middle third set? I couldn't figure out how we get the last inequality, if anyone can show me step by step that would be really appreciated. I need to understand this part. Or can anyone show me how can I get the lower estimate by using probability measure.
In the fourth line from the bottom it is established that if $3^{-(j+1)}\le|U_i|$, then $U_i$ intersects at most $2^j3^s|U_i|^s$ level $j$ intervals. If $j$ is so large that $3^{-(j+1)}\le|U_i|$ for all $U_i$, then each $U_i$ intersects at most $2^j3^s|U_i|^s$ level $j$ intervals. This means that $\bigcup_iU_i$ can intersect at most $\sum_i2^j3^s|U_i|^s$ level $j$ intervals altogether, but we know that in fact $\bigcup_iU_i$ intersects all $2^j$ level $j$ intervals, so $\sum_i2^j3^s|U_i|^s$ must be at least $2^j$, i.e.,
$$2^j\le\sum_i2^j3^s|U_i|^s\,.$$
The Cantor trinary set is defined by starting with the interval [0, 1].
Remove the middle third, leaving [0, 1/3]U [2/3, 1].
Remove the middle third of each of those intervals, leaving [0, 1/9]U[2/9, 1/3]U[2/3, 7/9]U[8/9, 1].
Remove the middle third of each of those and so on.
The first interval has length 1, the intervals in the second case have total length 2/3, the intervals in the third case have total length 4/9, etc.  After the nth step we will have 2^n intervals each with length 3^n so total length (2/3)^n.  As n goes to 0 that goes to 0.
That was treating each interval, and so the limiting set, as having dimension 1.  To get a non-zero length, treat them as having dimension &quot;d&quot;.
Each would then have &quot;length&quot; (3^d)^n= 3^(nd).  We would still have 2^n intervals so a total &quot;length&quot; 2^n/3^(nd)= (2/3^d)^n.  If 2/3^d is larger than one, that will go to infinity.  If 2/3^d is less than one, it will go to 0.  To have a finite, non-zero result, 2/3^d must be one.
2/3^d= 1 so 2= 3^d.  log(2)= log(3^d)= d log(3) so d= log(2)/log(3).
Whatever base you use for the logarithm that will be 0.6309....