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.

Complements of closed sets

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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