The Hilbert Cube $H$ is defined to be $[0,1]^{\mathbb{N}}$, i.e., a countable product of unit intervals, topologized with the product topology.

Now, I've read that the Hilbert Cube is homogeneous. That is, given two points $p, q\in H$, there is a homeomorphism $f:H\rightarrow H$ with $f(p)=q$.

What's confusing to me is that it seems like there seems to be a stratification of points. That is, there are

  1. Points contained in $(0,1)^{\mathbb{N}}$

  2. Points which have precisely $n$ coordinate a $0$ or $1$ for n a fixed natural number.

  3. Point which have countably many coordinates equaling $0$ or $1$ and countably many not and

  4. Points which have n many coordinates NOT equal to $0$ or $1$.

Now, for fixed $p$ and $q$ both in class $1$ or $3$ (or fix an n and use class $2$ or $4$), it's clear to me that there is a homeomorphism taking $p$ to $q$, simply by swapping around factors and using the fact that $(0,1)$ is clearly homogeneous.

But what are the homeomorphisms which mix the classes? In particular, what homemorphism takes $(0,0,0,\ldots )$ to $(1/2, 1/2,1/2,\ldots )$?

Said another way, for any natural number $n>1$, $[0,1]^n$ is NOT homogeneous, precisely because of these boundary points. What allows you to deal with the boundary points in the infinite product case?

As always, feel free to retag, and thanks in advance!

Edit In the off chance that someone stumbles across this question, I just wanted to provide a rough idea of the answer, as garnered from the link Pete provided in his answer.

If one has a point of the form $(1,p)$ in $[0,1] \times [0,1]$, then there is a self homeomorphism of $[0,1]\times[0,1]$ taking $(1,p)$ to $(q,1)$ with $q\neq 0, 1$. For example, one can use a "square rotation". From here, the idea is simple: given a point in $H$ of the form $(1, p_2, p_3, p_4,\ldots )$, apply the square rotation on the first two factors to get a new point of the form $(q_1, 1, p_2, p_3,\ldots )$. Now, apply the square rotation on the second two factors to get a new point of the form $(q_1, q_2, 1, p_3,\ldots )$. The point is that after $k$ iterations, the first $k$ coordinates are all in the interior.

Now one proves a techinical lemma that states that the infinite composition of these homeomorphisms is a well defined homeomorphism. The infinite composition maps the point $(1, p_2, \ldots )$ to a point of the form $(q_1, q_2,\ldots )$ which lies on the "interior" of $H$. Finally, using the fact that $(0,1)$ is clearly homogeneous, one can easily map $(q_1, q_2,\ldots )$ to $(1/2,1/2,\ldots )$.

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    $\begingroup$ I hesitated to say it originally, but since I am surprised that this interesting question has so few upvotes: I think this question would play on MO, and perhaps get more attention from true topology experts. $\endgroup$ – Pete L. Clark Aug 12 '10 at 3:53
  • $\begingroup$ @Pete. I debated about which of the two to post on. However, since I'm not a topologist (at this point, I'm a grad student researching certain kinds of compact Lie group actions on compact Lie groups), I wasn't able to gauge the difficulty of the question - and it's not research related. That said, I'm satisfied by the answers I've received here (I wish I could select both!), so I think I'll refrain from posting on MO. $\endgroup$ – Jason DeVito Aug 12 '10 at 5:41
  • $\begingroup$ Very nice explanation. I wonder if there is also an easy way for how a "corner", e.g. $(1,1,1,...)$ can be mapped into an inner point. $\endgroup$ – M. Winter Sep 25 '17 at 19:33
  • $\begingroup$ @M. Winter: You work 2 coordinates at a time. The argument I listed works no matter what $p$ is, even if you're at a corner. Even in 2 dimensions, there is already a "rotation" which maps a corner point to an edge point. $\endgroup$ – Jason DeVito Sep 25 '17 at 20:37
  • $\begingroup$ @JasonDeVito I see, thank you! $\endgroup$ – M. Winter Sep 25 '17 at 20:39

When I first saw the question, I intended to post the following link as an answer, but then I saw that Qiaochu had come first with an essentially equivalent answer. I still don't see anything lacking in his answer (I was the first to upvote it), but if someone is looking for a different reference, here is one:


  • $\begingroup$ This was exactly what I was looking for. Thanks! (If I could accept both, I would!). $\endgroup$ – Jason DeVito Aug 12 '10 at 6:07
  • $\begingroup$ Pete, the pdf on that site is no longer available. $\endgroup$ – Bombyx mori Dec 28 '12 at 23:14

In Jan van Mill's books (infinite dimensional topology, a prerequisite; the infinite-dimensional topology of function spaces) there are complete self-contained proofs (starting from basic stuff about separable metric spaces) [in essence the same proof in both, but the second book is more recent so might be more readily available]. It is based on Anderson's proof of this, and uses the Inductive Convergence criterion, which states in essence that when (for compact metrisable $X$) $h_n$ is a sequence of homeomorphisms of $X$, there is a sequence of $\epsilon_n > 0$ such that if $d(h_n, 1_X) < \epsilon_n$, where $1_X$ is the identity on $X$, and $d$ is a complete distance on the space of homeomorphisms of $X$, that then the limit of $h_n \circ h_{n-1} \circ \ldots h_1$ as $n$ tends to infinity exists and is a homeomorphism of $X$. Also, each $\epsilon_n$ depends only on the $h_i$ with $i < n$. This allows one to construct homeomorphisms of compact metrisable spaces in a very controlled way. Note that a metric on the Hilbert cube is such that a movement in a high coordinate only contributes a very small amount to the distance, so we can push problems to higher and higher coordinates and "solve" them there by a sufficiently small movement. With this criterion the proof is just 2 pages or so.


This appears to be a difficult result. It was first proven in 1931 by O. H. Keller (reference), but the paper is in German. Keller actually proves a stronger result: any infinite-dimensional compact convex set in $\ell^2$ is homeomorphic to the Hilbert cube. (I guess there is an obvious choice of such a set which is clearly homogeneous, but I can't think of one at the moment.)

There is another proof of the theorem in this paper of Ageev, which is in English.

  • $\begingroup$ If you don't mind, I'm going to hold off on accepting this (but I did vote it up). Unfortunately, I don't read German, and the second paper doesn't seem to directly address my weaker question. Since I'm also unsure why the stronger result implies the weaker result, I'm going to wait in the hope that someone can fill in the gap. Thank you! $\endgroup$ – Jason DeVito Aug 12 '10 at 0:26
  • $\begingroup$ Here is what I suspect: I think a product of countably many copies of the circle is homeomorphic to the Hilbert cube. Maybe it even embeds into l^2 as a compact convex subset. $\endgroup$ – Qiaochu Yuan Aug 12 '10 at 0:29
  • $\begingroup$ @Charles. I think you're right, but I don't see how it helps. Would you mind elaborating a bit? I can see that l^2 is homogeneous, but I'm not sure how that implies that H is... $\endgroup$ – Jason DeVito Aug 12 '10 at 2:17
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    $\begingroup$ @Qiaochu - A countable product of circles can't be contractible as it has nontrivial pi_1, or am I missing something? $\endgroup$ – Jason DeVito Aug 12 '10 at 2:35
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    $\begingroup$ I just stumbled on this question, so I thought I'd point out that a countable product of $S^1$'s is not contractible. As Pierre points out, it is homeomorphic to $S^1\times \Pi_\omega S^1$, and so if we let $\pi$ denote the fundamental group $\pi\cong \mathbb Z\oplus \pi$. So it is not trivial. $\endgroup$ – Cheerful Parsnip Mar 6 '11 at 18:45

In the meantime, an elementary and self-contained proof of homogeneity of the Hilbert cube was given in

The Homogeneous Property of the Hilbert Cube, by Denise M. Halverson, David G. Wright, 2012.

  • $\begingroup$ This is roughly based on the idea I sketched in my answer, indeed (Inductive convergence). $\endgroup$ – Henno Brandsma Nov 22 '17 at 16:18

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