# You won’t believe what this space is homeomorphic to

I’m currently reading A comprehensive introduction to differential geometry, a book by Michael Spivak, which I find extraordinarly clear and pleasurable to read. It is one of those rare books you can read as you were reading a novel, which flows smoothly and takes your mind along. Really good.

Anyway, in Spivak’s book the following exercise was proposed:

Let $C \subset \mathbb R \subset \mathbb R^2$ be the Cantor set. Show that $\mathbb R^2 \setminus C$ is homeomorphic to the surface:

We’ll call it $T$. We recall that the Cantor set is the “limit” of the following iteration: starting from the unit interval, delete the middle third of each connected component. So the first three iterations are:

We’ll call it $C$. Remember that the Cantor set is comprised only of the points we get at the end, and indeed it is made of points and not thick lines like in figure. So $\mathbb R^2 \setminus C$ is a plane with a peculiar set of holes cut in it (we can assume $C$ to be embedded in the line $y=0$, but feel free to choose your favorite embedding).

Anyway, it’s easy to spot the similarity between these two structures. It is as if $T$ represented the process of making $C$. This is somewhat true. First, let’s shrink $\mathbb R^2$ to an open disk around our chosen embedding of $C$. They’re homeomorphic, so no big deal. Yet this helps us with the visualization.

Let’s start from the whole open disk and remove the points of the first iteration of the Cantor set: once we make some room around the holes, we get

Notice the boundaries are not part of our surface, since we started with an open disk (hence no outer border) and the inner boundaries were victims of the piercing too. We then invoke the power of the third dimension and start to deform our space as to lift the holes from the plane we were stuck in until now, at the same time shrinking the outer border as to get a more symmetrical shape. We get the following surface, called a pair of pants:

Voilà! Notice again how this surface describes the first iteration in the construction of the Cantor set: from the unit interval (= one big hole) we remove the middle third and get two smaller intervals (= two holes). Since the construction is somehwat recursive, this means that to get to the second iteration, we can just glue the waist of two more pair of pants to the cuffs of the one we already got.

Can you see where this is going? Exactly to $T$! I already find this amazing, yet something even greater is lurking just behind the corner.

We know the plane and the sphere are really close topologically speaking, they differ just by a single point. What I mean is $\displaystyle K(\mathbb R^2) \cong S^2$

where $K(-)$ denotes the Alexandrov compactification of a space. That the isomorphism above holds is best seen on the other way around: if we remove a point from $S^2$ we can then widen the puncture to a big hole, flatten everything out, and we remain with an open disk, which, as we’ve already seen, is just a scaled-down version of $\mathbb R^2$. The Alexandrov compactification of $\mathbb R^2$ is the opposite procedure: we add a point to the plane, and we wrap everything up to get to a sphere (imagine reversing the previous procedure until you get to the punctured sphere: adding the new point fills the puncture).

But then $\displaystyle S^2 \setminus C \cong K(\mathbb R^2) \setminus C \cong K(\mathbb R^2 \setminus C) \cong K(T)$

So compactifying $T$ should give us a sphere with a Cantor set removed. Let’s see this: take $T$ and close the top hole (the waist of the first pair of pants we used) with a disk. This corresponds to adding the “missing point” of $\mathbb R^2$, therefore what we get is indeed a sphere minus a Cantor set. The second and third equivalences are apparent.

To get a whole sphere now, we need to close the “ends”, which has to be done carefully in order to be sure we’re doing sensible things. The first problem is, they’re out of reach: we need to bring them back from infinity (stacking pants brought us very far!). To do this, imagine shrinking each “level” of $T$, for example by making each level exactly half as big as the previous one. Then we know the total height of $T$ will now be finite: so we have all those ends at reach, finally! Even better, the ends have become just punctures, thus the Cantor set perfectly falls in place to fill them. In the end we get the equivalence $\displaystyle S^2 \cong \bar{K(T)}$

Another way to see this is to “open” the tree by moving the two main branches of $K(T)$ on opposite sides. Then closing the ends by rescaling, as we did before, we get a branchy surface which, however, is closed and simply connected (think about a “finite” version of it to convince yourself), hence homeomorphic to a sphere.

But what if we moved the branches while we do this? I mean, there’s nothing that forces us to keep $\check{T}$ laid out straight, we could curl it as to tangle the ends on the right with the ends of the left. Actually, if we’re careful enough, we can tangle them in such a way as to form a link! See this:

We happened to have built Alexander’s horned sphere:

This pathological (and slightly eerie) space is a famous counterexample to a generalization of Jordan-Schoenflies’ theorem:

The complement of any simple, non-degenerate loop on the sphere has exactly two connected components, both homeomorphic to a disk.

Which is the (apparently) pretty obvious statement that cutting a sphere in half will give you two (deformed) disks. But when you translate “simple, non-degenerate loop” to the equivalent “an embedding of $S^1$, the statement begs to be generalized:

The complement of any embedding of $S^{n-1}$ into $S^n$ has exactly two connected components, both homeomorphic to $B^n$.

Yet the horned sphere shows this generalization is doomed to fail: while its complement has indeed two connected components, the unbounded one (the exterior) is not homeomorphic to the three-dimensional ball. The reason is that we sucessfully tangled the two ends of $\check{T}$ as to make them inextricable: if you wind a loop around one “horn” (a branch of $\check{T}$), there’s no way to make it slip pass the tangle:

Then we cannot hope to shrink it to a point, thus we’ve shown the exterior of the horned sphere to not be simply connected, while $B^3$ is.

I actually learned about Alexander’s horned sphere’s role as counterexample in Topology class two years ago, yet the teacher didn’t say why it was a counterxample. So when today I realized the connection to Problem 17, I was deeply pleased. And as I never would have imagined such a simple, conceptual proof, I found it worthy of a post here. I hope you enjoyed it as well!

## Notes

 An homeomorphism, or isomorphism of topological spaces, is a bicontinuous transformation, i.e. a continuous deformation whose inverse is also continuous. This means: no cuts/rips, no gluing, no “complete thinning out” (while a rod is contractible to a segment, the contraction can’t be inverted so it is not an homeomorphism, it’s instead an isotopy).

 In mathematical lingo, an (n-)sphere is the boundary of an (n+1-)ball, which is instead inteded to be solid. To clarify: an orange is a 3-ball, while its peel is 2-sphere.

 Because it’s an infinite tree, this is the same thing as gluing $T$ to a copy of itself along the waist, which, interestingly, corresponds to making a sphere by gluing two open disks along their boundary.

 It is indeed one of those very frustrating situations in which something looking very simple is actually quite tricky to prove. The difficulty is given by the fact that a continuous curve can be nonetheless very pathological: jagged, curled up, non-rectifiable, whatever. Assuming the loop to be piecewise linear or smooth, the statement becomes indeed quite easy to prove. Yet continuity alone is a very short lever to push on.

 This is the Jordan part of the Jordan-Schoenflies theorem, which succeeds to generalize to higher dimensions.

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