# Wild Arcs with simply connected complements: two examples.

We start with this Fox-Artin example:

As we saw in this previous post: this arc contains just one wild point; hence its complement is simply connected. But we can say more than that: this arc $K$ is “cellular”: that means that there is a collection of smooth 3-balls $B_1 \supset B_2 \supset B_3....$ where $\cap^{\infty}_{i=1}B_i = K$. That isn’t too hard so see: start by taking a small round 3-ball that encloses the wild point and taking a union of that ball with a tubular neighborhood of the rest of the arc outside of the ball. That first gadget is “almost” a ball; it might have a few handles which can be filled in. Then take a smaller ball around the wild point and repeat the process with a smaller tubular neighborhood…and repeat.

In fact, we can do this with ANY arc that has just one wild point.

Therefore $S^3 - K = S^3 - \cap^{\infty}_{i=1}B_i = \cup^{\infty}_{i=1}(S^3 - B_i)$ which is homeomorphic to $R^3$; hence this wild arc complement is homeomorphic to the complement of a tame arc.
Aside: this shows that arcs are NOT determined by their complements whereas smooth (or p. l. ) knots are.

But this arc is still wild; it has penetration index 3 near the wild point.

Now consider this example

This is like the wild Fox-Artin arc talked about in the previous post in this series, except that “one stitch has been missed. It’s complement is simply connected; the proof of this fact is suggested by this diagram:

The small loop represents a generator of the fundamental group; note how it bounds a disk in the complement of the arc.

However the complement is NOT $R^3$! There are different ways to see that; here is a fun one:

This shows $K$ with a smooth simple closed curve $J$ in its complement. Note: $J$ is compact (it is a smooth knot). Therefore there are two different ways to use $J$ to show that $S^3 - K$ is NOT homeomorphic to $R^3$:

1. In $R^3$, every compact set lies inside of some ball $B$. But one can show that there is no ball $B$ that contains $J$ and misses $K$
2. One can use algebraic topology: the fundamental group of $(S^3 -K)-J$ is NOT finitely generated; in $R^3$, the complement of any smooth knot is finitely generated.

This gives an example of an open, simply connected manifold that is not homemorphic to $R^3$.

The interested reader can consult the following references for more detail:

R. Daverman and G. Venema: Embeddings in Manifolds, American Mathematics Society Graduate Studies in Mathemtics, Vol. 106, 2009: Section 2.8
T. B. Rushing, Topological Embeddings, Academic Press, (Pure and Applied Mathematics, Volume 52) 1973, Section 2.4.

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