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I am studying fractals autodidactically (I have never had any topology discipline) and according to Mandelbrot: a fractal has a Hausdorff dimension that exceeds the topological dimension. I understood what Hausdorff's dimension is: it is the value of $s$ (unique) for which a jump between infinity and zero occurs: $$ \dim_H(X):=\sup\{s:\mathcal{H}^s(F)=\infty\}:=\inf\{s:\mathcal{H}^s(F)=0\} $$ However, I have been trying to understand what the topological dimension is for several days and three concepts always appear to me: the small inductive dimension, the large inductive dimension and the lebesgue covering dimension. I do not understand the definition and the difference between the three and to understand the concept of fratal I just need to understand the concept of topological dimension. Could you explain the concept to me intuitively and with examples and then the mathematical formalism, please?

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    Xander nciely explained the intuitions in his answer. The technical details are tricky indeed, but a nice fact is that for separable emtric spaces (so anywhere "fractals" are usually defined like $\Bbb C$) all three functions agree in value, and also agree that the value for $\Bbb R^n$ equals $n$. Which was the whole point of defining these kinds of dimension functions in the first place: in the face of space filling curves etc. people wanted a real topological proof that $\Bbb R^m \simeq \Bbb R^n$ iff $n=m$. Brouwer was the first to fill in the details correctly, using yet another function – Henno Brandsma Feb 16 '21 at 22:50
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    [cont] he called "Dimensionsgrad" and which coincides in value with the three you mention for locally compact, locally path-connected sep. metric spaces. The Hausdorff dimension is not a topological invariant in that puttting an equivalent metric on the space can change its value. So topologists aren't a big fan, it's more of a geometric invariant, and so the definition you quoted is "hybrid": a mix of topology and geometry. – Henno Brandsma Feb 16 '21 at 22:56

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First off, the term "fractal" does not have a universally agreed upon meaning. I have written about this before on Math SE, so I won't go into that again. I will point out that Mandelbrot walked back from the definition you provide in the second edition of The Fractal Geometry of Nature, so even he was not entirely wed to that notion.

Inductive Dimensions

Regarding the meat of your question, you cite three different topological notions of dimension: the small inductive dimension, the large inductive dimension, and the Lebesgue covering dimension. For the sake of completeness, let us recall the definitions (note: this presentation assumes familiarity with some of the basics of point-set topology, i.e. the definitions of open and closed sets, limit points, boundaries, etc—these definitions are cribbed from James Robinson's Dimensions, Embeddings, and Attractors):

Definition: The small inductive dimension, denoted $\DeclareMathOperator{ind}{ind}\ind(\cdot)$, is defined as follows:

  1. $\ind(\varnothing) = -1$,
  2. $\ind(X) \le n$ if for every point $p \in X$, $p$ has "arbitrarily small" neighborhoods $U$ with $\ind(\partial U) \le n-1$, where $\partial U$ denotes the boundary of $U$, and
  3. $\ind(X) = n$ if $\ind(X) \le n$ but it is not true that $\ind(X) \le n-1$.

The large inductive dimension, denoted $\DeclareMathOperator{Ind}{Ind}\Ind(\cdot)$ is defined similarly, with (2) replaced by

  1. $\Ind(X) \le n$ if for every closed set $A \subseteq X$ and each open set $V \subseteq X$ which contains $A$ there exists an open set $U \subseteq X$ such that $$ A\subseteq \overline{U} \subseteq V \qquad\text{and}\qquad \Ind(\partial U) \le n-1. $$

Both of these definitions are pretty gnarly, but they are trying to get at the same essential idea: we can understand the dimension of a set by looking at the boundaries of subsets; the dimension of a space should be greater than the dimension of the boundary of an open subset of the space.

Thinking in a very Euclidean way, an open set in $\mathbb{R}^2$ is (more or less) a disk. The boundary of a disk is a circle, so $\mathbb{R}^2$ should be one dimension greater than a circle. An open set in the circle is (roughly) an interval, and the boundary of an interval consists of two points. Thus the circle should have dimension one greater than a set containing two points. The boundary of a discrete set is empty, and so a collection of discrete points should have dimension one greater than the emptyset. Thus

\begin{align} \DeclareMathOperator{Dim}{dim} \Dim(\mathbb{R}^2) &= 1 + \Dim(\text{a circle}) \\ &= 1 + (1 + \Dim(\text{two points})) \\ &= 1 + (1 + (1 + \Dim(\varnothing))) \\ &= 1 + (1 + (1 + -1))) \\ &= 2, \end{align} where $\Dim$ denotes some appropriate notion of dimension.

The precise definitions of the small and large inductive dimensions try to capture this idea in a more general way, with slight variations in the way in which the open sets of concern are chosen. As is often the case, the jump from intuitive to precise requires a great deal of work.

Covering Dimension

The Lebesgue covering dimension approaches things slightly differently:

Definition: The order of an open covering of a topological space is the largest $n$ such that there exist $n+1$ sets in the covering with non-empty intersection. An open covering $\mathscr{V}$ is a refinement of an open covering $\mathscr{U}$ if every member of $\mathscr{V}$ is contained in some element of $\mathscr{U}$.

A set $A$ has Lebesgue covering dimension less than or equal to $n$, denoted $\Dim_{L}(A) \le n$, if every covering has a refinement of order less than or equal to $n$. $\dim_{L}(A) = n$ if $\dim_{L}(A) \le n$, but not $\Dim_{L}(A) \le n-1$.

Again, there is a pretty intuitive observation about Euclidean space running around behind the scenes. Imagine trying to cover $\mathbb{R}$ by open intervals, and then throwing away as many sets as you can. What you will likely end up with is a collection of intervals which overlap "just a little" at their ends, see the figure.

enter image description here

No matter how gnarly the initial cover is, I can always find a way to "throw away" enough sets so that the remaining sets don't overlap too much. Specifically, no point on the real line will be contained in more than two of the covering intervals.

By contrast, if you attempt to cover the plane $\mathbb{R}^2$ by disks such that no point is contained in more than two disks, you will rapidly find that this is impossible—there must be some points contained in three disks.

Further Reading

Robinson, James C., Dimensions, embeddings, and attractors, Cambridge Tracts in Mathematics 186. Cambridge: Cambridge University Press (ISBN 978-0-521-89805-8/hbk). xii, 205 p. (2011). ZBL1222.37004.

Engelking, Ryszard, Dimension theory. A revised and enlarged translation of ”Teoria wymiaru”, Warszawa 1977, by the author, North-Holland Mathematical Library. Vol. 19. Amsterdam, Oxford, New York: North-Holland Publishing Company. Warszawa: PWN - Polish Scientific Publishers. X, 314p. $ 44.50; Dfl. 100.00 (1978). ZBL0401.54029.

Hurewicz, W.; Wallman, H., Dimension theory., 165 p. Princeton University Press (1941). ZBL67.1092.03.

  • Thank you for your help. Sorry for the delay in the response, but I have been researching some sites to try to understand some details in your response. His answer was very complete, especially I understood better thanks to his examples. However, I had some doubts. Could you help me, please? I noticed the Lebesgue covering dimension, but I didn't notice the difference between the small inductive dimension and the large inductive dimension. What is the difference between the small inductive dimension and the large inductive dimension in intuitive and mathematical terms? – Carmen González Feb 27 '21 at 02:24
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    @CarmenGonzález The difference is in the second condition. In a metric space, the second condition for the small inductive dimension says that small open balls around a point have boundaries of dimension smaller than the ambient dimension of the space. The large inductive dimension says that every closed set is contained in an open set with boundary of dimension smaller than the ambient dimension. As I said above, the conditions are fairly technical and come from a similar intuition: the boundary of a ball should be of smaller dimension than the ambient space. – Xander Henderson Feb 27 '21 at 02:49
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    In many commonly seen spaces, the small and large inductive dimensions will coincide. I believe that all you need for the dimensions to coincide is for the space to be separable, which is a relatively mild condition. I seem to recall that Engelking proves this result, but I don't have access to that book at the moment. In any event, I would argue that the distinction really isn't all that important, unless you are interested in those kinds of topological questions. – Xander Henderson Feb 27 '21 at 02:55
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    As your question is tagged [tag:fractals], and most interesting properties of "fractals" are not topologically invariant, you probably don't need to care about the fine distinctions between the topological notions of dimension. – Xander Henderson Feb 27 '21 at 02:56
  • Sorry, I didn't understand the definition. I am not a mathematician and did not have any topology discipline. I am learning self-taught little by little, but I really needed to understand the difference between the two definitions. Do you think you could add an explanation of the conditions for the small and large inductive dimension, in your repply, please? I needed to understand where the conditions / properties of each definition come from and what is the difference between them, but I have already searched the internet and there is little information, unfortunately. – Carmen González Feb 27 '21 at 03:06
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    @CarmenGonzález If you don't have a background in topology, then these very technical definitions are likely to go over your head. If you are self-studying and you really want to understand these definitions, I would suggest that you pick up a text on point-set topology (Munkres is a reasonable choice), and try to work your way through it. Pay special attention to the sections on separation axioms. It is unlikely that I can do much to help (topological dimension theory is not really my area, and I don't have a ton of examples ready to go). – Xander Henderson Feb 27 '21 at 03:24