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Let $\Omega \subset \mathbb R^2$ be a bounded $C^2$ domain. Is $\Omega$ then finitely connected? As I learned recently a domain in $\mathbb R^2$ is finitely connected iff “[its] complement has finitely many components (holes)“.

The first (mostly smooth) bounded domain with infinitely many holes that comes to mind is $$\Omega = B_1(0) \setminus \overline{\bigcup_{n=2}^\infty B_{1/n^2}(1/n)},$$ however this domain fails to be $C^2$ at $0$. (Here $B_{r}(x)$ is the ball with radius $r$ centered at $x$.)

Similar constructions should run into the same issue. In particular, suppose that $\Omega$ is not finitely connected and enumerate the holes by $A_1, A_2, \dots$. For $n \in \mathbb N$, take $x_n \in \partial A_n$. As $\partial \Omega$ is compact, there exists a converging subsequence. I doubt that $\Omega$ can be $C^2$ at this limit point. However, I fail to make this precise.

My main interest is to apply a result holding for finitely connected $C^2$ domains to a bounded domain and thus I would welcome a reference stating that bounded $C^2$ domains are always finitely connected. This is also the reason I require the domains to be $C^2$ but the same question can of course also be asked for Lipschitz domains, for instance.

Edit: A domain is $C^2$ if it can locally be written as the set of points above the graph of a $C^2$ function. A domain is bounded if it is contained in $B_R(0)$ for some $R > 0$. A bounded $C^2$ domain is a domain which is both bounded and $C^2$.

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    $\begingroup$ Could you define a "bounded $C^2$ domain"? $\endgroup$
    – Igor Belegradek
    Commented Aug 24, 2022 at 15:35
  • $\begingroup$ "it can locally be written as [...]" -- ? $\endgroup$
    – Wlod AA
    Commented Aug 25, 2022 at 8:29
  • $\begingroup$ @IgorBelegradek It is a bounded domain and its boundary is locally a graph of a $C^2$ function, in other words, the boundary is a $C^2$ submanifold of $R^2$. $\endgroup$
    – Piotr Hajlasz
    Commented Aug 25, 2022 at 12:53
  • $\begingroup$ @PiotrHajlasz: it is a terminology issue, i.e., what is the standard terminology in complex analysis? A bounded domain is usually defined as a bounded open connected set. When we say that a domain is $C^2$ we could mean a number of different things, e.g. that its closure is a (compact) $C^2$ submanifold of $\mathbb R^2$. With this definition the unit disk about the origin with the standard Cantor set removed is a bounded $C^2$ domain. The way the domain is defined in the edit this example works. $\endgroup$
    – Igor Belegradek
    Commented Aug 25, 2022 at 13:05
  • $\begingroup$ Here is one reasonable definition (not sure if it is standard). A bounded $C^2$ domain $D$ is a bounded open connected subset whose topological boundary (=$\bar D\setminus D$) is locally the graph of a $C^2$ function. Another approach would be to start with a compact $C^2$ submanifold of the plane and remove a closed subset from its boundary (e.g. remove a Cantor set from the boundary of the unit disk); what is left is a nice object, a (possibly noncompact and with boundary) $C^2$ submanifold of the plane. Both situations clearly deserve attention. Which one is standard? $\endgroup$
    – Igor Belegradek
    Commented Aug 25, 2022 at 13:59

2 Answers 2

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Yes. It is finitely connected. The boundary of your domain is a compact one-dimensional manifold and therefore it consists of a finite number of curves diffeomorphic to circles. You can find a proof of classification of compact one dimensional manifolds in:

J.W. Milnor, Topology from the differentiable viewpoint. Based on notes by David W. Weaver University Press of Virginia, Charlottesville, Va. 1965.

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  • $\begingroup$ Ah, that makes a lot of sense, thanks! And the point is that my concrete example above is not a smooth two-dimensional manifold so that we cannot conclude that its boundary is a compact (smooth) one-dimensional manifold, yes? $\endgroup$
    – Keba
    Commented Aug 25, 2022 at 8:18
  • $\begingroup$ @Keba You are right, not smooth at zero. The boundary of your domain is not a smooth submanifold of $R^2$, because the boundary in a neighborhood of zero is not diffeomorphic to an interval as it contains infinitely many circles. The boundary of a $C^2$ domain is however, a submanifold. $\endgroup$
    – Piotr Hajlasz
    Commented Aug 25, 2022 at 12:42
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I'm not sure what you mean precisely by bounded $C^2$ domain, but I'll take it to mean a compact set with positive reach (that includes slightly weaker regularity domains in fact).

Such domains are indeed finitely many connected. One way to see it is that, for $r$ less than the reach, the $r$-neighborhood of the domain deformation retracts to the domain, hence the inclusion of the domain into its $r$-neighborhood is an isomorphism at the homology level.

Since one can construct a polygonal domain sandwiched between the domain and its $r$-neighborhood, the inclusion map isomorphism in homology described above factors through a finite dimensional space, hence its rank, which is equal to the Betti numbers of the domain, is finite.

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    $\begingroup$ Thanks for this alternative answer. It seems to require more knowledge about homology theory (which I lack) and thus I have accepted the other, more elementary, answer. $\endgroup$
    – Keba
    Commented Aug 25, 2022 at 8:20

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