I am quite new to the forum so please feel free to correct/give pointers if I am posting something in the wrong place.

I have been perusing "Dimension Theory" by Witold Hurewicz & Henry Wallman and while doing so noticed I couldn't quite see the reason for 'studying/researching' dimension theory. As an undergraduate when I was taught the concept of dimension in terms of 'linear independence' and 'spanning' of vectors I never paid much heed to alternate/analytical formulations of dimension. Now having come across, say, the Hausdorff, Fractal, Assouad dimensions I am quite curious as to why mathematicians have focussed so much on this avenue.

I would be very, very grateful for any pointers or links which can give me a little bit of 'general' motivation for studying dimensions i.e why study dimension theory. Specific examples from dynamical systems, metric geometry, number theory, set theory et cetera are very welcome too.

Thank you, thanks a lot!

P.S the tags might be a little over the place - couldn't quite create new ones for metric geometry and assouad dimension.

  • $\begingroup$ Dimension is just a form of measurement. Don't you think it is informative to measure things? $\endgroup$
    – rschwieb
    Commented Mar 23, 2016 at 10:08
  • $\begingroup$ Oh yes, I understand that. I was merely wondering if there had been a big open problem which had drawn interest towards the study of dimensions. For instance, problems along the lines of calling two spaces the 'same' in the sense of homeomorphisms. For instance some of the interest in metric geometry and calculus on metric spaces inspired the Assouad dimension. Studying attractors of differential equations can be transformed into embedding problems utilising the notions of dimensions, et cetera. That's the kind of thing I meant... $\endgroup$
    – Amsar
    Commented Mar 23, 2016 at 12:00
  • $\begingroup$ If $I^n$ is a unit cube in $\mathbb{R}^n$, then we could ask if there is a function $d:I^n \rightarrow \mathbb{N}$, such that $d(I^m)=m$. This lead to the dimension thery. $\endgroup$
    – Hulkster
    Commented Mar 25, 2016 at 23:37

2 Answers 2


Size, contrary to the common adage, is very important. Measuring how large things are is a very elementary and useful mathematical practice. For instance, given a set $A$ which is a subset of $B$, if you can prove that the size of $B$ is strictly bigger than that of $A$, then you can conclude that there is something in $B$ which is not in $A$. This is something we do all the time. A striking example is Cantor's proof of the existence of transcendental numbers, where one can show the cardinality of the algebraic numbers is strictly smaller than the cardinality of the real numbers, and thus conclude that transcendental numbers exist. This example shows that it is useful to be able to measure how big things are even beyond the realm of discrete finite sets.

So, whenever you are presented with an object you should ask yourself "how big is it?". Sometimes the answer is obvious, but often it is not even clear how to measure the size of the given object. Also, different perspectives may dictate different notions of size. In any case, we measure how large things are since it helps us compare them to other things and reason about them. And that principle holds true for topological objects. Different ways of measuring the size of a geometric object may give different answers, and provide different properties, and thus teach us various things about the given object. Dimension theory in topology\geometry is the equivalent to counting. Both counting and dimension theory are important for similar reasons.

  • $\begingroup$ Thank you for the answer. I really like the reference to Cantor's proof. Hmm, I wonder what your opinion is as to why this 'intuitive' notion of measuring 'size' was not satisfied by the creation of 'measures'? I suppose the difficulty arises in trying to put a metric on something which is fundamentally topological, cue problems of metrizability? $\endgroup$
    – Amsar
    Commented Mar 22, 2016 at 9:57
  • $\begingroup$ measures 'count' side in a different way. The rationals for instance are countable, so are small relative to the reals, i.e., they have measure 0. But the closure of the rationals are the reals, so in that respect they are large with respect to the reals, i.e., they are dense. Size matters, and size may change according to perspective. $\endgroup$ Commented Mar 22, 2016 at 19:37

"Topological dimension" as defined by Hurewicz and Wallman is a topological invariant which lets you prove that many pairs of spaces are not homeomorphic to each other.

For example, why are $\mathbb{R}$ and $\mathbb{R}^2$ not homeomorphic?

The pedestrian answer is: $\mathbb{R}$ is separated by removal of any point, whereas $\mathbb{R}^2$ is not.

But you might want to generalize: Why are $\mathbb{R}^m$ and $\mathbb{R}^n$ not homeomorphic when $m \ne n$?

To obtain a general answer, return to the special case and recast it in terms of dimension theory:

  • $\mathbb{R}$ has dimension $1$ (because a point separates it, in fact every point locally separates, which is how dimension 1 is defined in some treatments; I believe the Hurewicz and Wallman definition is close to this but perhaps not exactly the same)
  • but $\mathbb{R}^2$ does not have dimension $1$.
  • and "dimension 1" is a topological invariant.

In general, topological dimension is an invariant assigned to topological spaces whose value is either a non-negative integer or $\infty$. Homeomorphic spaces must have equal topological dimension. Using it, the proof that $\mathbb{R}^n$ is not homeomorphic to $\mathbb{R}^m$ when $n \ne m$ is that the dimension of $\mathbb{R}^n$ equals $n$ whereas the dimension of $\mathbb{R}^m$ equals $m$ (these must be verified, of course).

You've mentioned a few other types of dimension in your question, and each of them is an important invariant in a different context. For example, fractal dimension or "Hausdorff dimension" which I presume you mean, is a real-valued bi-Lipschitz invariant of metric spaces. Hausdorff dimension is what I would use to prove that the middle thirds Cantor set is not bi-Lipschitz equivalent to the every-other-fifths Cantor set, because their Hausdorff dimensions are $\log(2)/\log(3)$ and $\log(3)/\log(5)$ respectively, and these numbers are not equal.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .