Note: I'm not too experienced with branches of mathematics like topology, so a lot of the terminology I use is used pretty loosely.

I think we all understand the fact that a point in $n$-dimensional space is described with $n$ parameters. In 2d space, you have $(p_1, p_2)$ (where $p_k \in \mathbb{R}$), in 3d space you have $(p_1, p_2, p_3)$, so in an $n$-dimensional space you have $(p_1, p_2, ..., p_n)$. I know the way I'm stating this with the notation I use isn't as robust as it could be, but you get the idea.

Going into the world of fractal geometry, we find non-integer dimensional spaces. Applying what I said previously, how would you even express a point in such a space? What does it even mean to be that kind of space? Are you unable to express a "point" in such a space?


Dimensions in Vector Spaces

The notion of dimension that you put forward is, roughly speaking, the correct way of understanding the dimension of a vector space. A vector space consists of two basic sets of objects: a base field $\mathbb{F}$, and a set of vectors $\mathcal{V}$. These two sets are linked via some algebraic structures:

  • $\mathbb{F}$ is a field, which means that it has two operations (addition and multiplication) which "get along" in the ways that we would expect ($\mathbb{F}$ is closed with respect to both operations; both operations are commutative and associative; multiplication distributes over addition; every element has an additive element; every nonzero element has a multiplicative inverse; etc). Elements of the base field are generally referred to as scalars.
  • $\mathcal{V}$ is a group, which means that it has an addition operation that behaves nicely. That is, $\mathcal{V}$ is closed with respect to the addition operation, addition is commutative and associative, and every element has an additive inverse. The elements of $\mathcal{V}$ are called vectors.
  • The vectors and scalars get along in the sense that if $\alpha$ is a scalar and $\vec{v}$ is a vector, then $\alpha \vec{v}$ is another vectors. This operation of "gluing" a scalar to a vector is called scalar multiplication. Scalar multiplication is also nicely behaved, in that it distributes over the addition of vectors (among other nice properties).

The basic example of a vector space that you should have in your head is $\mathbb{R}^n$ (vectors are tuples of real numbers, and the scalars are also real numbers). However, there are other examples, such as $\mathbb{C}^n$ (a complex vector space), $\mathbb{Q}_p^n$ (a $p$-adic vector space), and function spaces such as $C(\mathbb{R})$ and $L^p(\mathbb{R})$ (vectors are certain types of functions, and the scalars are real numbers).

Assuming the Axiom of Choice, it can be shown that every vector space has a basis. The basis of a vector space is a collection of vectors that span the vector space. In particular, we say that $\mathcal{B} = \{\vec{e}_i\}_{i=1}^{N}$ is a basis for the vector space $\mathcal{V}$, then we can write every element of $\mathcal{V}$ as a linear combinations of elements of $\mathcal{B}$. That is, if $\vec{v} \in \mathcal{V}$, then there is a collection of scalars $\{\alpha_i\}_{i=1}^{N}$ such that $$ \vec{v} = \sum_{i=1}^{N} \alpha_i \vec{e}_i. $$ Moreover, the elements of $\mathcal{B}$ are linearly independent, which means that if $$ \sum_{i=1}^{N} \alpha_i \vec{e}_i = 0, $$ then $\alpha_i = 0$ for all $i$. Even more impressively, while a vector space may have many different bases, every single one of them has the same cardinality. That is, if $\{\vec{e}_i\}_{i=1}^{M}$ and $\{\vec{f}_j\}_{j=1}^{N}$ are both bases for $\mathcal{V}$, then $M=N$ (note that we could have $M = N = \infty$—bases need not be finite; this is why we require AoC).

This is all a long explanation, but the punchline is that we can think of the vectors in a basis as describing the different "directions" in which it is possible to travel in a vector space. Each direction requires a different parameter to describe, which gives the dimension of the vector space. That is, the dimension of a vector space is the cardinality of any basis.

This is a purely algebraic notion, and has purely algebraic generalizations (such as the Krull dimension). However, this is not the kind of dimension that people who study fractals generally think about when they say "dimension."

Dimensions through Scalings

To talk about dimensions the way that most fractal geometers think about them, you need both more and less structure. On the one hand, we don't really need a vector space—typically, all that is required is a metric [measure] space. A metric is a way of determining the distance between two points, while a measure assigning a size (or volume) to subsets. We typically want both a metric (which we can use to describe how sets scale) and a measure (though we can make due without an a priori measure).

The basic intuition is as follows:

  1. Consider a line segment $L$ of length $\ell$. If we scale that line segment by a factor of $\alpha$, the length of the scaled segment will be $\alpha$. That is, $$\operatorname{length}(\alpha L) = \alpha \operatorname{length}(L).$$ Note that line segments have dimension 1 (we'll come back to this in a minute—remember that we are building intuition right now).
  2. Now consider a square $S$ with side length $\ell$. If we scale the square by a factor of $\alpha$, the area of the scaled square will be given by $$ \operatorname{area}(\alpha S) = (\alpha \ell)^2 = \alpha^2 \ell^2 = \alpha^2 \operatorname{area}(S)$$ (you should check to make sure that you believe this). Note that a square is two dimensional.
  3. Finally, consider a cube $C$ of side length $\ell$. Again scaling the cube by a factor of $\alpha$, we have $$ \operatorname{volume}(\alpha C) = (\alpha \ell)^2 = \alpha^3 \ell^3 = \alpha^3 \operatorname{volume}(C). $$ Note that a cube is three dimensional.

The pattern is as follows: let $X$ be an $n$-dimensional set with measure $\mu(X)$ (if $X$ is a segment, then $\mu(X)$ is the length of that segment; if $X$ is a cube, then $\mu(X)$ is the volume of that cube; and so on). Scale $X$ by a factor of $\alpha$. Then $$ \mu(\alpha X) = \alpha^n \mu(X). $$ The goal is to describe the dimension in terms of the measures of scaled sets, so we solve for $n$: $$ \mu(\alpha X) = \alpha^n \mu(X) \implies \alpha^n = \frac{\mu(\alpha X)}{\mu(X)} \implies n = \log_{\alpha} \left( \frac{\mu(\alpha X)}{\mu(X)} \right). $$ We can use this to define a notion of dimension in terms of scaled copies of sets and measures. Roughly speaking, $\frac{\mu(\alpha X)}{\mu(X)}$ is the number of copies of $X$ are required to "cover" $\alpha X$. If we take the logarithm of this quantity, we get the dimension.

For example, consider the ternary Cantor set $\mathscr{C}$. This set is constructed by removing the middle third of an interval, then removing the middle third of each of the remaining intervals, and so on. Wikipedia gives the first several steps in the construction:

enter image description here

Notice that below the initial interval, the left- and right-halves of the Cantor set look the same. Indeed, the Cantor set is made up of two identical copies of itself, scaled down by a factor of $\frac{1}{3}$. In the language developed above, we have $$ \frac{\mu\left(\frac{1}{3} \mathscr{C} \right)}{\mu(\mathscr{C})} = \frac{1}{2} \implies \dim(\mathscr{C}) = \log_{\frac{1}{3}} \left( \frac{1}{2} \right) = \log_3(2). $$ This we can reasonably state that the dimension of the Cantor set is $\log_3(2) \approx 0.631$.

Note, however, that the notion of dimension here is quite different from the notion of a vector space. Indeed, while the ternary Cantor set lives in $\mathbb{R}$ (a vector space), the Cantor set itself doesn't have an obvious vector space structure. Instead, it has a way of measuring distances between points (the absolute value from $\mathbb{R}$), and a rough way of measuring the sizes of subsets (there is actually something very subtle going on here, related to something called the Hausdorff measure, which is a way of getting a measure from a metric). In this context, the dimension is a way of describing how the measure of a set changes with scaling.

Concluding Thoughts

Typically, when we talk about spaces with non-integer dimension, we are thinking about something like the notion of dimension described above. That is, we have in mind some way of quantifying how the volumes or measures of sets change as the sets are scaled.

That being said, there are actually many other notions of dimension running around. There are purely topological definitions, such as the covering dimension, big and little inductive dimensions, and so on. The dimension of a manifold is an integer related to the local structure of that manifold. The previously mentioned notion of Krull dimension is algebraic. There are notions of dimension somewhat related to, but more general than, the idea of relating measures and scales. There are even ways of assigning complex valued dimensions to sets (this is my own area of research).

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  • $\begingroup$ Wonderful answer! Thanks! $\endgroup$ – Sam Dec 15 '17 at 16:45

There are there are several notions of dimension, one more general than the other:

  • The dimension of a vector space, e.g. for the vector space $\mathbb{R}^n$ it is the number $n$. This dimension has the property that it is the smallest number $n$ for which every point $x\in\mathbb{R}^n$ can be expressed as an $n-$tuple $(x_1,...,x_n)$.

Given any subset $S\subset\mathbb{R}^n$ and a point $x\in S$, you can still express it as an $n-$tuple, after all $x$ also lies in $\mathbb{R}^n$. However you may lose the property that $n$ is the smallest number for which you have an expression like that. E.g. consider $S= \{x=(x_1,...,x_n)\in\mathbb{R}^n \vert ~ x_n = 0\}$. It is evident that in order to describe points in $S$ you only need the first $n-1$ numbers. It turns out that $n-1$ is again the smallest such numbers, so you're inclined to call it the dimension of $S$.

Now consider the example $S= \{x=(x_1,x_2) \in \mathbb{R}^2 \vert ~x_1^2+x_2^2 = 1\}$, which is just a circle. Again, any point $x\in S$ may be described by two real numbers, but again there is some redundancy, because it actually suffices to give the angle $\phi$ such that $(x_1,x_2) = (\cos(\phi),\sin(\phi))$ to specify a point. You need at least one such coordinate to specify points, so $1$ is the smallest number of necessary coordinates. Here you would like to say the $S$ has dimension $1$.

  • The dimension of a manifold (something that locally looks like $\mathbb{R}^n$, certainly fractals are not manifolds) generalizes the concept from the two examples. It is the smallest number of coordinates which suffice to specify points.

Vectors spaces are also manifolds and in this case the two definitions of dimension coincide. Now for any arbitrary subset $S\subset \mathbb{R}^n$, which is not as nice as a manifold, think of your favourite fractal, you can still define a dimension, e.g. by using the following notion:

  • The Hausdorff-dimension. Its definition is completely independent of the previous thoughts and at first glance its not at all clear why it is a generalization.

However, if $S$ is a manifold then its dimension coincides with its Hausdorff-dimension. Otherwise the dimension of $S$ might very well be non integral, for example the Cantor set $C\subset \mathbb{R}$ has dimension $\log_32$. You already see that it does not make any sense to speak of $\log_32$ as the least number of coordinates and indeed the notion of dimension simply loses its characterization in terms of minimal number of coordinates, if your set is ill behaved. But as said at the very beginning, $S$ remains to be a subset of $\mathbb{R}$, so points in $C$ can still be expressed by $1$ coordinate. You could also consider the set $S = C \times \{0\} \subset \mathbb{R}^2$. This is essentially the same as the Cantor set and has the same dimension. S, being a subset of $\mathbb{R}^2$ can be described using two coordinates, however $1$ coordinate suffices, as the second will be redundand. But this all does not have anything to do with the dimension anymore.

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