Let $E \subset \mathbb{C}$ be compact and totally disconnected. Is there an elementary way to prove that $\mathbb{C} \setminus E$ is connected?

  • 5
    $\begingroup$ A closely related question on MathOverflow: mathoverflow.net/questions/55718/… $\endgroup$ Commented Apr 8, 2011 at 2:48
  • 2
    $\begingroup$ Indeed, it's a nice question, but I have no reason to think that there is a better answer available than the one given there by William Thurston (who is, to say the least, a very eminent mathematician). Is the OP satisfied with this answer? $\endgroup$ Commented Apr 8, 2011 at 3:25
  • $\begingroup$ I didn't know there was such a question on mathoverflow. The answer given by Thurston is indeed very nice, but it seems to rely on non-trivial homology / cohomology results. I was looking for an easier proof, perhaps one that uses specific properties of the complex plane and might only works in that specific case. $\endgroup$ Commented Apr 8, 2011 at 15:33

1 Answer 1


There are many "elementary" topological ways of proving this. Of course at some point you will have to use some properties of the complex plane. For example, it is not difficult to show that any set that separates the Riemann sphere must contain the boundary of a domain $U$ whose complement is connected; i.e. $U$ is simply-connected, and hence $\partial U$ is connected. So any set (not necessarily closed, by the way) that disconnects the plane contains a plane continuum that disconnects the plane. Of course this still uses something (as mentioned in the question on MathOverflow, a key fact is that the union of two disjoint compact sets that do not disconnect the plane also does not disconnect the plane).

Here is the most elementary proof that I can come up with on the spot. Again, we will need to use some properties of the plane, and we will also use elementary properties of Hausdorff convergence of continua.

Lemma. Suppose that $A\subset\mathbb{R}^2\setminus\{0\}$ disconnects $0$ from $\infty$. Then $A$ contains a nontrivial plane continuum that separates $0$ from $\infty$.

Proof. First we note that, for some $\varepsilon$, $A$ contains a closed set $$B\subset \{x\in A: \varepsilon < \|x\| < 1/\varepsilon\}$$ that separates $0$ from $\infty$. Indeed, this follows from the definition of connectedness: There is an open, bounded set $U$ such that $\partial U$ is contained in $A$, so we can set $B := \partial U$.

Now cover the plane by a grid of small squares of sidelength, say, $1/n$ ($n$ sufficiently large), and let $B_n$ be the union of (closed) squares that intersect $B$. Then $B_n$ separates $0$ from $\infty$. Now it is an elementary (though perhaps tedious) exercise to show that $B_n$ has a connected component $\tilde{B_n}$ that separates $0$ from $\infty$. (This is essentially a discrete fact about a two-dimensional grid.)

Let $C$ be a Hausdorff limit of the sequence $\tilde{B_n}$. Since each $\tilde{B_n}$ has diameter greater than $2\varepsilon$, and this sequence is uniformly bounded, the limit is a nontrivial continuum. By construction, $C\subset B$, and it is not difficult to see that $C$ must also separate $0$ from $\infty$ (though we do not need this for the original question).

Does this makes sense or am I missing something?


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .