Intuitive picture for topological and fractal dimensions of a fractal The usual one-liner definition given for what a fractal dimension represents, is along the lines of: a measure of the change in complexity of the fractal pattern at different scales of measurement. 
On the other hand, for the topological dimension of a fractal, it is often described as a union of open covers/sets that contain the topological space in which the fractal lies.
These two definitions are both very vague and confusing to me, likely because I sadly lack a good understanding of topology in general, but I was hoping maybe someone can elucidate the meaning behind these two quantities with simple examples of fractals. 
 A: There are multiple definitions (not always equivalent) of topological dimension and of whatever you mean by "fractal dimension". Let's stick to metric spaces, without going to general topological spaces. 
Topological dimension
The inductive definition of topological dimension says: the space is $0$-dimensional if it can be covered by arbitrarily small open subsets with empty boundary. For example, the middle-third Cantor set qualifies because it can be covered by $2^n$ intervals, each of size $3^{-n}$ (as small as we wish), and each of them has empty boundary (is both open and closed).  Two more observations: 


*

*a finite metric space is $0$ dimensional (just cover it by one-point subsets); 

*a connected metric space is not $0$-dimensional (unless it's one point): there is no way to write it as the union of disjoint open sets, by the very definition of "connected".  


Then, a space is $1$-dimensional if it's not $0$-dimensional but can be covered by arbitrarily small open subsets with $0$-dimensional boundary. The von Koch snowflake (or any simple curve for that matter) is $1$-dimensional: on one hand, it's connected, so cannot be $0$-dimensional; on another, it can be covered by tiny subarcs, whose boundary has $2$ points, so is $0$-dimensional.  Two more examples: 


*

*Cartesian product of the Cantor set with a line segment is $1$-dimensional. 

*The Sierpinski carpet is $1$-dimensional: cut it into small pieces with $0$-dimensional boundary by using vertical and horizontal lines that pass through centers of removed squares.  


The topological dimension is always an integer, by definition. So it neglects a good deal of information about the fractal structure; there is no difference between a line segment and a snowflake curve. 
Hausdorff/Minkowski/box/packing dimensions, whatever you call them
Ignoring the details which are responsible for the variety of names, let's say that a $d$-dimensional metric space is one where we need about $\epsilon^{-d}$ sets of size $\epsilon$ to cover it, when $\epsilon$ is small. So, a square is two-dimensional: use an $n\times n$ grid of smaller squares, with $\sqrt{2}/n \le \epsilon$. The Cantor set has $d=\log 2/\log 3 $ because we need $2^n$ subsets of size $3^{-n}$, so choosing $n$ so that $3^{-n}<\epsilon$ yields about $\epsilon^{-d}$ sets. The Sierpinski carpet requires $8^n$ smaller squares of size $3^{-n}$, so its dimension works out to $\log 8/\log 3$, between $1$ and $2$. 
A: Let consider any Cube $Q\subset\mathbb R^d$ with edges of length $c$ Then 
denote by $2Q$ the corresponding cube obtain by doubling all edges of $Q$. 
We not that $Vol(2Q) = 2^d vol(Q)$. in fact if $l*Q$ denote the cube obtained by multiplying all edges of $Q$ by $l$ then 
we have 
$vol(Q)= l^d vol(Q)$. This formula holds true also for euclidean balls and some oder nice geometric object. See that the topology of $\mathbb R^d$ is completly defined cubes or Balls
So intuitively a family of objects $(F_i)_i$ are called to be fractals of dimension $\alpha$ if there generate a topology in which the following holds true 
$$ vol(l*F_i) =l^\alpha vol(F_i).$$
