If you roll out a lump of clay into a snake, then tie a string loosely around it, can you contort the ends of the snake, without actually pressing them together, so that you cannot get the string off?
You can stretch the clay into a Medusa’s head of snakes, and tangle them as you like, again without letting them touch. If you are allowed to rest the string on the surface of the clay, then you can get it off: you just slide it around and over what was an end of the original snake.
If however no point of the string may touch the snake, then you can trap the string so that it cannot be removed. You let the clay start as an amphisbaena, with heads at both ends; you let each head be a hydra’s head, sprouting two new ones from the stump when cut off; and you cut off all heads as they appear, ad infinitum. The result is a solid binary tree, as discussed in “The Tree of Life.” With the branches of the tree intertwined just so, you get what is called Alexander’s Horned Sphere.
You may object that you cannot actually do something ad infinitum. A person called James R. Meyer has this objection. His website is devoted to showing what’s wrong with mathematics and philosophy. I too sometimes take issue with contemporary professional philosophy in my blog. I like to think that I am saved from being a crank by knowing, as a mathematician, that the truth cannot be given uniquely to me. Meyer himself has a page about cranks, taking issue with how others define them. I found first his page about the horned sphere. The page shows no sign of understanding what the horned sphere is for.
The horned sphere is a topological sphere. It arises as a counterexample to a formerly conjectured threedimensional version of the Jordan–Schönflies Theorem.
I read about that theorem as a child, or at least about the simpler form, the Jordan Curve Theorem: a simple closed curve divides the plane into an inside and an outside. I did not understand what there was to be excited about. I suppose now the point is the following. A simple closed curve is a continuous function f from the unit interval [0, 1] into ℝ^{2} that repeats a value precisely at 0 and 1, so that, f being the function,
f(t) = f(u) ⇔ t = u ∨ {t,u} = {0, 1}.
A simple example is g, given by
g(t) = (sin (2πt), cos (2πt)),
tracing out the unit circle; but f might not be given by a formula. How are you going to define a function h on ℝ^{2} so that h(x,y) = 1, if (x,y) is “inside” the curve given by f, and otherwise h(x,y) = 0? I just did define h, but only by begging the question of what inside means.
If g is as above, and f is some simple closed curve, then f∘g^{−1} is a homeomorphism between the two curves, just for being continuous in both directions. According to the Jordan–Schönflies Theorem, the homeomorphism extends to a homeomorphism from the whole plane ℝ^{2} to itself.
Passing to three dimensions, one may think that if there is a homeomorphism from a sphere, considered as a surface in ℝ^{3}, to some other surface in space, then that homeomorphism should extend to a homeomorphism from the whole space ℝ^{3} to itself.
The horned sphere shows that one would be wrong. Here I want to work out some details of the proof. It may serve as another example of my recent theme, that mathematics is the science whose findings are proved by deduction. Topology in particular can seem to be a counterexample, since it seems to rely on physical intuition, albeit an intuition that tolerates supposed absurdities like completed infinite processes.
The horned sphere has been the inspiration of some sculptures pictured in Claire Ferguson, Helaman Ferguson: Mathematics in Stone and Bronze (Erie, Penn.: Meridian Creative Group, 1994). A friend recently gave me the book, and the book is a reason for this post.
J. W. Alexander described his construction in “An example of a simply connected surface bounding a region which is not simply connected” (Proc. N. A. S. 10, 1924). A simply connected space is one in which

you can carry a string from any point to any other like Ariadne, and

if you carry the string back to where you started from, then you can hold the two ends and draw the whole string to yourself.
In more technical language, the loop of string must be nullhomotopic, meaning there is a continuous function f from the square [0, 1] × [0, 1] into the space in question such that the function t ↦ f(0, t) is the original loop, and t ↦ f(1, t) is constant.
Alexander describes his construction with words and a drawing:
The surface Σ is the limiting surface approached by the sequence Σ_{1}, Σ_{2}, Σ_{3}, .. It will be seen without difficulty that the interior of the limiting surface Σ is simply connected, and that the surface itself is of genus zero and without singularities, though a hasty glance at the surface might lead one to doubt this last statement. The exterior R of Σ is not simply connected, however, for a simple closed curve in R differing but little from the boundary of one of the cells γ_{i} cannot be deformed to a point within R.
The surface Σ is the surface of our clay. To say that it is of genus zero means it has zero holes; a torus has genus one. I myself do see without difficulty that Σ will have no holes. I am not sure what Alexander means by singularities. That the exterior of Σ is not simply connected is not clear without a proof. Alexander himself confesses, at the end of his short article,
This example shows that a proof of the generalized Schönfliess theorem announced by me two years ago, but never published, is erroneous.
If he was wrong then, why is he not wrong now?
The relevant Wikipedia article, “Alexander Horned Sphere,” lists a reference on my shelves, Spivak’s Comprehensive Introduction to Differential Geometry, Volume One (2nd ed., Houston: Publish or Perish, 1979). Defining Alexander’s Σ using a drawing like his, an exercise asks the reader to show what I said was clear, that Σ has no holes. (Spivak calls the surface S and asks, “Show that S is homeomorphic to S^{2},” the latter being the surface, which is twodimensional, of a sphere.) The exercise then asserts, without explicitly asking for a proof, that the S together with the the region outside it fails to be a manifoldwithboundary, though the student has shown that it would have to be one, if S were differentiable.
I pass to another of the Wikipedia references, Hocking and Young, Topology (Reading, Mass.: AddisonWesley, 1961), which says on page 175 (where the circumflex on ĥ is a tilde in the original),
Let S be a simple closed surface in [Euclidean space] E^{3}, that is, S is a homeomorph of S^{2}, and let h be a homeomorphism of S onto the unit sphere S^{2} in E^{3}. Is there an extension ĥ of h such that ĥ is a homeomorphism of E^{3} onto itself? … Alexander … gave a famous example, the Alexander horned sphere, showing that the answer must be “no” in the general case. This example is pictured [below]. We can see from the picture alone that it is quite obvious that the complement of the horned sphere is not simply connected. Since the complement of S^{2} in E^{3} is simply connected, it follows that no homeomorphism of E^{3} onto itself will throw the horned sphere onto S^{2}. Note that there is a Cantor set of “bad” points on the horned sphere.
My sense is that in topology nothing is obvious. In Real Analysis II in graduate school, the professor made an “obvious” topological assertion that I questioned and later disproved with the help of a book I found: Steen and Seebach, Counterexamples in Topology. I am going to have to work out a proof of what Hocking and Young think is obvious. I am helped by yet another reference in the Wikipedia article, though a reference to which the link was dead, till I revived it: Allen Hatcher, Algebraic Topology.
We started with a lump of clay and rolled it out into a snake. Alternatively, we pulled two horns from it. Now continuing, from each horn we pull two more horns, ad infinitum.
More precisely, letting the initial lump of clay be B_{0}, we pull two horns out of it to get B_{1}. In general, B_{n} will have 2^{n} horns, and when we make each horn into two, we get B_{n+1}. Each B_{n} is closed and bounded, hence compact.
Perhaps ordinary clay is incompressible, so that our original lump retains its volume throughout the reshaping. We prefer the volume to grow, as if we add horns at each step. Let us say that B_{n} has 2^{n} horns, one for each binary sequence σ of length n. Let the corresponding horn itself be called H_{σ}. To obtain B_{n+1}, to each H_{σ} we attach the horns H_{σ0} and H_{σ1}. Now we have a strictly increasing chain:
B_{0} ⊂ B_{1} ⊂ B_{2} ⊂ …
We may think of B_{n} as a balloon filled with air; when we lower the pressure outside, the balloon expands from B_{n} into B_{n+1}.
Since it depends on physical intuition, this description is perhaps not “universally valid.” We could write down equations in three variables, defining the surfaces of the B_{n}. Then inequalities would define the solids themselves.
There is a homeomorphism h_{n} from B_{n} to B_{n+1}. We “could” define it precisely, but we don’t want to bother. Still there are conditions it must satisfy. For each n, we have that the composite function
h_{n}∘ … ∘h_{0}
is a homeomorphism from B_{0} to B_{n+1}. We want the sequence
(h_{0}, h_{1}∘h_{0}, h_{2}∘h_{1}∘h_{0}, …)
of these functions to converge uniformly to an injective function h. This is not automatic; we have to choose the h_{n} right. The uniform convergence will imply that h itself is continuous. Every closed subset of B_{0} is bounded and therefore compact, so its image under h is compact and therefore closed. Thus the inverse of h is also continuous, so h is a homeomorphism onto its image.
We call that image B. Now this is homeomorphic to the original B_{0}. Why then have we bothered to create B? We want the complement of B in space not to be homeomorphic to the complement of B_{0}. We achieve this by attaching in two steps to H_{σ} the horns H_{σ0} and H_{σ1}, for each binary sequence σ of length n, for each n:

Attach a “handle” to H_{σ}.

Cut out part of the handle, leaving the horns H_{σ0} and H_{σ1}.
The cutout part will include the handles that will be attached to the horns H_{σ0} and H_{σ1} in the next step. Thus B_{n} gets 2^{n} handles attached, resulting in X_{n}. These form a decreasing chain of compact sets. In particular, we now have
B_{0} ⊂ B_{1} ⊂ B_{2} ⊂ … ⊂ B ⊂ … ⊂ X_{2} ⊂ X_{1} ⊂ X_{0}.
It is not automatic that B is the intersection of the X_{n}, but we can have ensured that this will be so.
The key move really needs three dimensions, and the pictures just above don’t handle this. For each n, for each σ of length n, the handles attached respectively to the horns H_{σ0} and H_{σ1} should be interlocking. This is to ensure that

the complement of X_{n+1} is not simply connected,

every loop in the complement of X_{n} that is nullhomotopic in the complement of X_{n+1} was already nullhomotopic in the complement of X_{n}.
If one accepts this, then the complement of B also fails to be simply connected; for, by the compactness of X_{0}, any loop in the complement of B that is nullhomotopic must already have been nullhomotopic in the complement of one of the X_{n}, and we know that that complement has loops that are not nullhomotopic. Indeed, suppose we are given a homotopy of loops in the complement of B, namely a continuous function f from the square [0, 1] × [0, 1] into the complement of B such that f(0,0) = f(0,1) and f(1,0) = f(1,1). Let K be the image of the square under f. Then K ∩ B is empty, but is the intersection of the descending chain of closed subsets K ∩ X_{n} of the compact set X_{0}. Therefore, by compactness, some set in the chain must be empty. Thus f is a homotopy in the complement of some X_{n+1}.
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