Consider the following situation. Suppose we have a closed oriented Riemannian surface $ \Sigma $ and a connected open subset $ \Omega \subseteq \Sigma $ with a boundary, consisting of finitely many smooth arcs. Consider a disk $ D $ centered at a point $x \in \Omega $ and denote by $ \Omega_0 $ the connected component of $ \Omega \cap D $ that contains $ x $.

I wish to relate the number of connected components of $ \partial \Omega $ to the number of connected components of $ \partial \Omega_0 $. Is it true that the first is not less than the latter?

Any comments and references are welcome. Thanks!

  • $\begingroup$ Why should this be true? For $D$ sufficiently small, $\partial\Omega_0$ has exactly one connected component, no matter what $\Omega$ is. $\endgroup$ – Alex Degtyarev Mar 7 '15 at 8:45
  • $\begingroup$ Sorry, I was not clear - by a "small" disk, I didn't mean "arbitrary small". $\endgroup$ – Boggie Georgiev Mar 7 '15 at 11:37

(Based solely on the topology:)

Since $\Omega_0$ is connected and contained within the disc $D$, it must be a planar surface, ie. a disc with holes. It has one 'outer' boundary component that is made up at least in part of $\partial D$ (unless $\partial D$ is disjoint from $\Omega$, in which case $\Omega_0=\Omega$), and all other boundary components of $\Omega_0$ are complete boundary components of $\Omega$. Thus $\Omega_0$ could have one more boundary component than $\Omega$ (when $\partial D$ is contained in the interior of $\Omega$) and otherwise it has at most as many as $\Omega$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.