TITLE: Is the property of having connected complement inherited by the closure of components of interior?
QUESTION [8 upvotes]: Let $E$ be a compact set in the plane whose complement $\Omega$ is connected. Let $A$ is the interior of $E$, $B$ is some component of $A$. Can I assert the complement of the closure of $B$ is connected?

In other words: if a compact set does not disconnect the plane, then neither do the closures of the components of its interior. This is easy to see for the components themselves, but the interior-closure steps complicate the problem.

REPLY [3 votes]: This statement is actually false.
The counterexample, known as the "Lakes of Wada" is apparently a classical result of Yoneyama (1917).
See this Cut The Knot article, or A First Course in Algebraic Topology by Czes Kosniowski.
Essentially, in this example the sea water is $E^c = \Omega$. The cold water, warm water and dry land together form $E$. They are bounded by the sea, and together are closed - hence they are a compact set, with a connected complement.
The dry land is actually the common boundary of all three open sets - the sea water ($E^c$), cold water and warm water!
The interior $A$ is thus just the cold water and warm water - which are two disjoint open sets. If we take the cold water as $B$, then the closure of the cold water component is just the cold water together with the dry land, and the complement is of course the sea water with the warm water.
But again, these are just two disjoint open sets. Therefore the complement of the closure of the component is disconnected.
Wow!