5
$\begingroup$

Let $\mathscr{C}$ be the class of compact subsets of the (euclidean) plane $\mathbb R^2$ with connected complement. If $K\in\mathscr C$ and $M \subseteq \mathbb R^2$ is homeomorphic to $K$, does it follow that $M$ belongs to $\mathscr C$?

(The corresponding question for open subsets (and the complement in the extended plane) has a positive answer because simple connectedness is a topological property.) The question is related to complex analysis (e.g. Megelyan's theorem).

$\endgroup$
5
  • $\begingroup$ Aren't your sets still simply connected? $\endgroup$
    – Hagen von Eitzen
    Jun 2, 2015 at 12:43
  • 1
    $\begingroup$ @HagenvonEitzen: If you define simple connectivity in the standard way that every loop is homotopic to a point, then no. The main reason is that compact connected sets might not be path connected. E.g., if you close a $\sin 1/x$ curve, then it disconnects the plane but every loop in it is contractible. $\endgroup$
    – Lukas Geyer
    Jun 2, 2015 at 13:02
  • 1
    $\begingroup$ For reasonably nice sets (locally contractible) this is true by Alexander duality, but I am not sure about the general case. $\endgroup$
    – Lukas Geyer
    Jun 2, 2015 at 13:03
  • 2
    $\begingroup$ @LukasGeyer: There is a version of alexander duality that uses Cech (co)homology instead, and has no conditions on the sets. Because 0th Cech homology counts quasicomponents, but because the complement of our compact set is open (hence locally connected), this is the same thing as counting components. So Alexander duality suffices. $\endgroup$
    – user98602
    Jun 2, 2015 at 15:55
  • $\begingroup$ @MikeMiller If I understand your comment well this answers the question. If you formulate it as an answer I would like to accept it. $\endgroup$
    – Jochen
    Jun 3, 2015 at 13:31

1 Answer 1

Reset to default
3
$\begingroup$

We can answer this with Cech cohomology. There is a theorem known as Alexander duality, stated in full generality as follows:

Let $A$ be a compact subset of $S^n$. Then $\check {H}^{n-q-1}(A;\Bbb Z) \cong H_q(S^n - A; \Bbb Z)$. (I think Bredon proves this in his book "Geometry and Topology".) All groups here are understood to be reduced.

Your question is: let $A$ be compact. Is "number of path components of $S^n-A$" independent of the embedding into $S^n$? And the answer to this is yes: $H_0(S^n - A)$ is precisely $\Bbb Z^{a-1}$, where $a$ is the number of path components of $S^n - A$. So we see that the number of path-components of $S^n - A$ is one more than the rank of $\check H^{n-q-1}(A;\Bbb Z)$, which is a homeomorphism invariant.

Lastly, note that $S^n - A$ is an open subset of $\Bbb R^n$, and hence path-components are the same as connected components, because it's locally path-connected.

It is going to be very difficult in general to calculate the Cech cohomology of $A$. If your compact set $A$ is locally contractible, then this is the standard singular cohomology, which is much easier to compute.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .