Why do we care about the Hilbert Cube?

Question

The Hilbert Cube is defined to be the countable infinite Cartesian products of the interval $[0,1]$ or anything homeomorphic to $[0,1]$. Why do we care about this object?

I appreciate questions like these, but I also think that every salad needs a little vinegar, so I'll just say it. I have a Ph.D. in math, I had to use a good amount of topology in my thesis, not too much. And I personally do not care about the Hilbert cube. It's just one of those "yeah, there's lots of weird stuff in topology" things to me. You could spend your life inside examples like that but I don't know why anyone would. It's OK if it is also like that for you. I would not try to discourage anyone from learning more about it. There could be good ideas in its construction or what it shows. — leslie townes, Feb 08 '21 at 01:52
@leslietownes wow thats interesting. but does the hilbert cube has applications? I know the cantor set is another object that shows up in some chaotic dynamical systems. Is that the case for the Hilbert Cube? — willyx888, Feb 08 '21 at 02:14
My own view, which is blinkered by prejudices, is that a lot of topological curiosities mainly have application in the sense that they remind you that you need to pay attention to topology, and nothing is automatically what you'd expect from working with $\mathbb{R}^n$ or Lie groups or whatever all the time. And despite my somewhat bitter prior comment, I am thankful to those examples for that. It keeps me nervous about using topology. But to answer your question I have not seen "real life" examples, whether in applications or pure math, that turn upon the Hilbert cube. — leslie townes, Feb 08 '21 at 02:17
The Hilbert cube is not just a "topological curiosity". Look up "Chapman's Theorem", which plays an important role in the classification of manifolds and cell complexes. — Lee Mosher, Feb 08 '21 at 02:42
@leslietownes correct me if im wrong, but aren't there many theorem that requires the hilbert cube to be proven? Such as the Urysohns Theorem. Or maybe that is just the text I am using. — willyx888, Feb 08 '21 at 02:45
In my view, yes, exactly. In my own view of the mathematical world, and for my purposes, examples like this primarily exist to illustrate the logical distinctions between various collections of axioms/assumptions, and not for their own use in proving things. But I may be wrong. I've certainly seen theorems whose proofs become trivial if you can show that whatever objects the theorem is talking about can be embedded in something that I would personally probably regard as needlessly complicated. But I don't know of examples of this type with the Hilbert cube. It sounds as though others do. — leslie townes, Feb 08 '21 at 03:52

Henno Brandsma · Accepted Answer · 2021-02-08T09:13:13.123

Every compact metrisable space (of any dimension ) is homeomorphic to a closed subset of the Hilbert cube. Any separable metrisable space is homeomorphic to a subspace of the Hilbert cube. Every separable metrisable topological vector space is homeomorphic to the pseudo-interior $(0,1)^{\Bbb N} \simeq \Bbb R^{\Bbb N}$ of the Hilbert cube.

There is a very nice theory of Z-sets and homeomorphisms of the Hilbert cube (see van Mill’s books on infinite-dimensional topology, or Bessaga and Pelczynski’s book for more on these theorems). It’s a fundamental object in infinite-dimensional topology. The hyperspace of any Peano continuum (in the Hausdorff metric) is homeomorphic to it too, e.g.

score 2 · Answer 2 · answered Feb 08 '21 at 17:53

In addition to Henno's answer and many comments, here is my take, as a geometric topologist. Below is a quote from the introduction to Chapman's book "Lectures on Hilbert cube manifolds" (Q-manifolds, which are analogues of topological manifolds, except that the local structure is modeled on the Hilbert cube):

Why do we care about the Hilbert Cube?

2 Answers2