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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!

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  • $\begingroup$ 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
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(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 $\Omega$.

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