# How to see the Whitehead continum is a Cantor set?

In the construction of Whitehead manifold, a 3-manifold, open noncompact and contractible but not homeomorphic to $\mathbb R^3$, Whitehead used a set of nested tori. I can understand the construction in this way, but how to see the limit set is a Cantor set? What is the homeomorphism from the limit set to the standard Cantor set? and what is the Hausdorff dimension of this set?

• I am not familiar with the construction itself, but a continuum is never a Cantor set. And in general, you show that something is a Cantor set not by finding an explicit homeomorphism to the standard Cantor set, but by showing that it is compact, totally disconnected, and perfect (i.e., has no isolated points). Commented Nov 26, 2013 at 15:32
• Limit set obtained in the process is not Cantor. What made you think it is? Commented Nov 26, 2013 at 19:47
• It is Whitehead continuum, not continuum. You can take a look at wiki's page en.wikipedia.org/wiki/Whitehead_manifold
– Sun
Commented Nov 26, 2013 at 19:48
• @studiosus, I think it is Cantor set times an interval.
– Sun
Commented Nov 26, 2013 at 19:55
• This limit set is connected, unlike product of Cantir set and the interval. Commented Nov 26, 2013 at 20:32

## 2 Answers

As others said, the Whitehead continuum $W$ is a continuum: a compact connected set. (Relevant: Arbitrary intersection of closed, connected subsets of a compact space connected?) Hence, it's not homeomorphic to a Cantor set.

That said, the visual resemblance of $W$ to the Cantor set $C$ can be made precise: see the blog post On Whitehead-type manifolds by Conan Wu, where $W$ is glued from two copies of $C\times [0,1]$.

what is the Hausdorff dimension of this set?

To ask this question, you have to give the set a metric first. There is no canonical metric on $W$; if it's considered as a subset of $\mathbb R^3$, it will surely have the induced metric, but the formation of this subset is subject to arbitrary choices (how thick are the tori? how are they positioned? how exactly is the linkage shaped?). If we follow Wu's construction using the standard $1/3$ Cantor set, the resulting set has Hausdorff dimension $1+\log 2/\log 3$, which is the Hausdorff dimension of $C\times [0,1]$. But $C$ could just as well be another Cantor-type set, with dimension anywhere between $0$ and $1$.

Another reason why the Hausdorff dimension of a topological space isn't a well-defined concept: if $d$ is a metric on $X$, then so is $\sqrt{d}$, and the Hausdorff dimension of $(X,\sqrt{d})$ is twice the Hausdorff dimension of $(X,d)$.

The topological dimension of $W$ is $1$, same as for $C\times [0,1]$ and for the same reason: there is a basis in which the boundaries of basis elements are totally disconnected.

I saw at least two constructions of this Whitehead manifold.

Let me shorten solid torus into torus.

One presented in Gabai's course notes, where it is defined as the increasing union of tori $T_n$, with $T_n$ contained in $T_{n+1}$, each being immersed in the next one in the same knotted way.

The one on the Wikipedia page https://en.wikipedia.org/wiki/Whitehead_manifold (as of now) is as the complement in the 3-sphere $S^3$ of the intersection $K$ of decreasing nest of tori with $T_{n+1}$ contained in $T_n$ in the same knotted way as above (but note $n+1$ and $n$ have been permuted).

The two constructions give diffeomorphic manifolds.

Indeed recall the complement of a torus in $S^3$ is a torus. Passing to complements in $S^3$ in the first step $n=1$ in the first construction yields two tori as in the first step $n=1$ second construction. Same holds for other steps.

My opinion is that Sun was speaking of $K$.

There is much flexibility in the construction so even though the set $K$ has to be connected (and thus cannot be a Cantor set unlike, for instance, Antoine's Necklace), it may or may not have empty interior depending on choices, its fractal dimension could be anything $\geq 1$. One can arrange so that in most places (probably not all?) $K$ is locally a Cantor times an interval. One interesting approach is to try to do it with an IFS.