Topological vs Fractal

If you were to ask a physicist what a dimension means to him, he would likely explain that it refers to the number of degrees of freedom of motion that exist. However, in the first course of Linear Algebra, the concept of a linear space's dimension is introduced, which refers to the number of vectors that exist in the basis. We can relate this to the mechanical definition by saying that we can think of these degrees as a basis. $$\\$$ In a different sense, "What is the number of dimensions in a square?". I think most of people would say it has two dimensions, although it has a bounded space, so it is not really a linear space!. Hence, the first definition is limited to only the linear spaces. But what can we say more about dimensions?. $$\\$$

Why dimensions needed?

A square is considered to have two dimensions because it has length and width, but not height. Typically, these dimensions help us determine the size of a space and its relation to other objects in space. However, when discussing fractals, using only length and width dimensions would not provide any new information about these complex shapes. Therefore, mathematicians have developed new types of dimensions that can provide more insight into fractals. $$\\$$

Fractal Dimensions

$$\\$$

Hausdorff Dimension

I have intoduced the definition of Hausdorff dimension in my thesis (you can find my thesis posted in this blog) as following: $$\\$$ The Hausdorff Dimension for a set $A \subseteq R^{n}$ is defined by: \[H_{dim}(A)= \inf \{ s: H^{s}(A)=0 \} \] Hausdorff dimension is one of the fractal dimensions. But why we consider it as a fractal dimension, that can be answered by taking an example of a famous fractal: "Cantor set" $$\\$$

The middle third Cantor set

There is a more general fomrs of Cantor set, but the most famous one is the middle third Cantor set which I have talked about in my thesis. $$\\$$ $\textit{The middle third Cantor set}$ is a set of points on a line segment made by removing the middle third of some interval (say we have [0,1] as our interval, we remove $ (\frac{1}{3},\frac{2}{3})) $ then removing the third middle of both of intervals resulted in the previous step, and continuing inductively in an infinite number of steps. $$\\$$ Let $F$ denote for The middle third Cantor set. I have proved in my thesis that $dim(F)=\frac{\log(2)}{\log(3)}$. Actually, this number tells us about the formation of such an inductive-formed set. Let $F_{1}$ be the first third and $F_{2}$ be the last third. $F_{1}, F_{2}$ are both Borel sets. $$\\$$ $$H^{s}(F) = H^{s}(F_{1}) + H^{s}(F_{2})= (\frac{1}{3})^{s} H^{s}(F) + (\frac{1}{3})^{s} H^{s}(F) = \frac{2}{3^{s}} H^{s}(F) $$ If $0< H^{s}(F) < \infty$ then: $$s = \frac{\log(2)}{\log(3)}$$ As we can see, the number is not an integer. The calculations used to obtain it are closely related to the process used to construct the Cantor set. $$\\$$ We can make another Cantor set, where the middle third is of length $\frac{1}{2}$ and the other two equal thirds are of length $\frac{1}{4}$. We can calculate its dimension by the same way, and we can predict it to be $\frac{\log(2)}{\log(4)} = \frac{1}{2}$ We can say generally that, if the first and last third is of length $\alpha$, then the dimension of the obtained Cantor set $= \frac{\log(2)}{\log(\frac{1}{\alpha})}$ $$\\$$ So this number describes how the inductive process is going in some sense. $$\\$$ There are several definitions of fractal dimensions, such as Minkowski dimension and spherical dimension. However, they all share a common property which makes them "fractal" - they account for irregularities at very small scales. I plan to write an article on another fractal dimensions, but for now, I will go for the topological dimensions

Topological Dimensions

Topological dimensions refer to the dimensions that remain unchanged under homeomorphism. I will talk about "The Covering Dimension" as an example. $$\\$$

Covering Dimension

Preliminary Defintions

For two families of sets $A,B$ such that $A$ is an open cover of a metric space $S$, We say $B$ is a refinement for $A$ if and only if $B$ is also an open cover and for every $\beta \in B$ there exist $\alpha \in A$ such that $\beta \subseteq \alpha$ $$\\$$ A family $A$ of set is said to have order $\leq n$, if every $n+2$ sets of $A$ have empty intersections. This family has order $=n$ if it has order $\leq n$ but not $ \leq n-1$ $$\\$$

The main definition

For a metric space $S$, we say $A$ is of dimension $\leq n$ if and only if every open cover of $A$ has refinement of order $\leq n$. $A$ is of dimension $n$ if it is of dimension $\leq n$ but not of dimension $\leq n-1$ $$\\$$ You can see that this topological dimension takes only integer values. $$\\$$ We can also see that it is a topological property because it depends on the simplest intersections of an open cover. $$\\$$ The covering dimension for the Cantor set we have defined above is zero!. I will explain the reason to you, but not in a rigorous manner. $$\\$$ Covering dimension idea is about how the sets in the open cover treat each other. The Cantor set is compact nowhere-dense, which means we can always have a refinement of sets which its supremum and infimum is not in it!((inf,sup) $\cap F$), that mean we can always get a disjoint cover, that means any 2 sets of this cover is disjoint. From the definition above, It is $\leq 0$. no nonempty set can have negative dimension, so it is zero-dimension. $$\\$$ I think it is obvious that the information we got by the Hausdorff dimension is more detailed for fractals, as it takes care of the sets in small scales. Also we can see that the topological dimension here is $\leq$ fractal dimension, which we have stated in the last post.