Complements of closed sets

Construct  two homeomorphic closed subsets $A,B$ of some Euclidean space such that their complements are not homeomorphic, but show that if $A,B$ are closed homeomorphic subsets of $\Bbb R^{n+m}$ such that $A$ lies in $\Bbb R^n\times 0$ and $B$ lies in $0\times\Bbb R^m$, their complements in $\Bbb R^{n+m}$ are homeomorphic.

4 thoughts on “Complements of closed sets”

1. Take the unknot and the trefoil knot in $\mathbb{R}^3$. Both are obviously homeomorphic, but the knot group of the unknot (that is, the fundamental group of its complement) is isomorphic to $\mathbb{Z}$, while the knot group of the trefoil is isomorphic to $B_3$, the braid group on $3$ strands.

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1. It’s even possible to do this in $\Bbb{R}^2$. Take $A$ the union of the unit circle and the origin, $B$ the union of the unit circle and a point outside of it. The two are obviously homeomorphic but the complement of $B$ has a contractible connected component, while the complement of $A$ does not.

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2. To understand the following solution follow this steps (we will draw a simple example): Let $m = n = 1$. Draw in the horizontal axis $A = \{\text{a point and a closed interval}\}$. Draw in the vertical axis $B =\{\text{a closed interval and a point}\}$.
Now, we know that we have a homeomorphism $f : A \to B$. This map is in particular a bijection, use it to draw the points in $\mathbb{R}^2$ of the form $(a,f(a))$, call this set $G$ and notice that it also consists of a line and a point. Look at the picture, the claim is that the complement of $A$ is homeomorphic to the complement of $G$, which, by symmetry, is homeomorphic to the complement of $B$.
This is pretty credible: See how $G$ is the graphic of $f$, since $A$ is closed we can use Tietze’s theorem to extend $f$ to all $\mathbb{R}$ and then moving the graphic of this extension up and down we can cover all the plane.

Let us prove this formally, so now we are in the general $\mathbb{R}^{n+m}$ case. Observe that $f$ can be seen as a map $f : A \to \mathbb{R}^m$. Since $A$ is closed, by Tietze’s theorem there is a lift $\overline{f} : \mathbb{R}^n \to \mathbb{R}^m$. Define a map from $\mathbb{R}^{n+m}$ to $\mathbb{R}^{n+m}$ by
$(v,h)\mapsto (v,\overline{f}(v)+h)$, which is clearly continuous. It is easy to see that restricting and corestricting it, we get a continuous bijection between the complement of $A$ and the complement of $G = \{(a,f(a)) \mid a\in A\}$. Since it is injective, by invariance of domain, it must also be open and thus it is a homeomorphism.

This argument shows that a homeomorphism between $A$ and $B$ can be lifted to a homeomorphism of the intire space $\mathbb{R}^{n+m}$.

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1. You don’t need to appeal to invariance of domain, since $\bar f$ has an explicit inverse!

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