TITLE: complement of a totally disconnected closed set in the plane
QUESTION [23 upvotes]: While preparing a course in complex analysis, I stumbled over a remark in Dudziak's book on removable sets, namely that any totally disconnected $K \subset\subset {\mathbb C}$ must have a connected complement; a remark, "that verifying the reader may find one of those exercises in 'mere' point-set topology that is a wee bit frustrating". Out of curiosity I spent an evening with this question. The assertion turned out to be a simple consequence of Theorem 14.2 ("If $x$ and $y$ are separated by the closed set $F$ in the open or closed plane they are separated by a component of $F$.") in an old book "Elements of the Topology of Plane Sets of Points" by M.H.A. Newman, Cambridge 1951. There, the proof is based on an lemma by Alexander (used in his proof of the Jordan-Brouwer separation theorem; Trans. AMS 23, 333-349, 1922), stating that, for disjoint closed sets $F_1$ and $F_2$ in the plane, two points which are connected in the complement of $F_1$ and in the complement of $F_2$ are connected in the complement of $F_1 \cup F_2$. This lemma fails for more general surfaces (Newman gives a counterexample for the torus) and is proved by homological methods. So, here are my questions:
(a) is there a more modern reference to these kind of results (Newman's book uses quite an ideosyncratic terminology and notation);
(b) is there a simpler proof not using Alexander's lemma (or similarily deep results);
(c) how about the connectedness of the complement of a totally disconnected closed set in  surfaces (or topological spaces) more general than the plane?

REPLY [10 votes]: I know this is an old question that already has an excellent answer (although it may apply only to compact sets). However, let me respond to the original question, concerning the following results:

Lemma. If two points $\newcommand{\R}{\mathbb{R}}x,y\in\R^2$ are disconnected from each other by the closed set $F$, then they are disconnected by a connected component of $F$.

It is clear (as noted) that the specific topology of the plane has to be used in some way, as e.g. this is no longer true on the torus. The key is indeed the following. 

Janiszewski's Theorem. If two compact sets $A,B\subset S^2$ do not separate $S^2$, and $A\cap  B$ is connected, then $A\cup B$ also does not separate $S^2$.

This result goes back to Janiszewski (1912), rather than the article of Alexander you mention. It is key to proving many "obvious" facts in plane topology (in particular it can be used to give a topological proof of the Jordan curve theorem). For example, a quick search found these notes giving a direct proof of Janiczewski's theorem online. I also seem to recall that there is a referene to a topological proof in Pommerenke's "Boundary behaviour of conformal maps", but I don't have the book to hand at the moment.
In any case, it should hopefully be clear that a direct argument can be furnished (e.g. by modifying a curve connecting two given points by replacing certain chords with those of simple closed curves sufficiently close around the two sets), but it is always going to be a bit fiddly.
Remark 1. The result in the form you quoted:

Corollary. If two compact sets $A,B\subset S^2$ do not separate the points $x,y\in S^2$, and $A\cap  B$ is connected, then $A\cup B$ also does not separate $x$ and $y$.

follows easily from Janiczewski's theorem (filling in the components of $S^2\setminus A$ that do not contain  $x,y$, and similarly  for $B$).
Remark  2. A proof of the lemma above (which turns out to be very useful), in slightly different formulation, can be found e.g. in my paper Connected escaping sets of exponential maps, DOI:10.5186/aasfm.2011.3604 where it was given for completeness, although in that sketch the relevance of Janiczewski's theorem is brushed under the carpet in the last line.