Please, clarify relationship between Hausdorff dimension and storage space of points in Cantor sets. Why is it that points in a Cantor set do not require less storage space (encoding bits) than points in a Cantor set with larger Hausdorff dimension?
If 2 Cantor sets have different Hausdorff dimension, then encoding set's points (on a computer file) should require less storage space on the set with lower dimension, same as encoding points in the plane requires double space compared to encoding the same number of points on a line.
So, storage space should depend on the dimension of the set containing the points.
But each point on each Cantor set is equivalent to a point in the other set, so they should require the same storage space to be encoded.
The Cantor set has a tree structure, then encoding a specific point in the tree should not depend on the Hausdorff dimension. A binary string of bits is all that's needed (0 for left branching, and 1 for right branching).
The only difference meanwhile encoding the same point on different Cantor sets should be writing the dimension, but that data is only written once per set, and has no effect on the storage space of each individual point.

^On this image, storing all the 32 points in the base of the tree requires 32 strings of 5 bits each one, no matter what is the Hausdorff dimension.
If it were an image, I can see that the lower dimensioned set would occupy less percentage of pixels, but that's because information is lost under the pixel size. A similar case is when the coordinates are stored on integer format: on lower dimensions there are more blank spaces, but locating the points require higher precision to avoid losing information, so the storage space saved non encoding blank spaces, is compensated with the extra storage needed for precision.
(I'm a programmer, and not a mathematician, so please, avoid using too much symbolism if possible).
 A: In order to relate fractal dimension of a set to storage space required for the set, you'll also need to specify the accuracy to which you are working. For simplicity, let's suppose we are working to $n$ bit accuracy and not worry about the finer points of floating point arithmetic. Then, there are $2^n$ such numbers and we require $n$ bits to store each number for a total of $n\times 2^n$ bits of storage required.
Now, suppose we want to store Cantor's ternary set to the same accuracy. Of course, it's more natural to think of that set in base 3; in fact, none of those points have a finite binary expansion but, using a standard rounding procedure, we're looking for the $n$-bit binary numbers that are closest to the Cantor set. Well, two points in the $m^{\text{th}}$ level approximation to the Cantor set can be a minimum distance $3^{-m}$ apart. Thus, to determine how many points from our set of binary numbers we need, we must consider the inequality
$$3^{-m} < 2^{-n}.$$
Once $m$ is large enough to make this true we won't generate any new points in the Cantor set closer to our $n$-bit number set. Put another way, we only need $n\times 2^m$ bits to store the Cantor set, where $m$ is the smallest integer such that
$$m>n\frac{\log(2)}{\log(3)}.$$
We just so happen to see the fractal dimension of the Cantor set in this formula.

Note that the same considerations apply if we compare the storage space required for two self-similar Cantor sets. Suppose that such a set consists of two copies of itself scaled by the factor $r$ and we wish to store it to an accuracy of $n$ bits. We now need to consider the inequality
$$r^m < 2^{-n}.$$
As is well known, though, the dimension is related to this scaling factor by the formula
$$1/r = 2^{1/d}.$$
For concreteness sake, suppose we consider the storage required required for a Cantor set of dimension $d_1 = 1/2$ compared to a set of dimension $d_2 = 1/4$. The second set is smaller and we expect it to require less storage space. 
Note that the scaling factors for the two sets are different. For the first set it is $r_1 = 1/4$, since 
$$1/(1/4) = 4 = 2^{1/(1/2)}.$$
Thus, we're led to the inequality
$$1/4^m<2^{-n}$$
or $m>n/2$. Thus, we need to compute the tree to level $n/2$. For the second set, $r_1 = 1/16$ and the inequality
$$1/16^m<2^{-n}$$
leads to the solution $m>n/4$. Thus, we only need to compute the tree to level $n/4$.
