# A five-part problem that uses ends and the Cantor set to prove that there are $c$ non-homeomorphic connected open subsets of $\mathbb{R}^2$

My question comes from Spivak's "Comprehensive Introduction to Differential Geometry Vol 1" (It's Chapter 1, Problem 24).

Background: Let $X$ be a connected, locally connected, locally compact, and hemicompact Hausdorff space.

And end of $X$ is defined to be a function $e$ that assigns to each compact set $C$ of $X$ a connected component of $X-C$, in such a way that if $C \subset D$, then $e(D) \subset e(C)$.

Let $E(X)$ be the set of all ends of X. In a previous problem, I've shown that the set $X \cup E(X)$ can be given a topology with basis elements being the open sets of X together with sets $N(C,e)=e(C) \cup \{f\in E(X) | e(C)=f(C)\}$ for each end $e$ and compact set $C$. (Thanks to Henno Brandsma here for suggesting the hemicompactness condition that makes this work). This topology is compact and Hausdorff.

Problem: The problem has five parts (a through e). I'd like to find a proof of b,d, and e. I think I have a solution for a and c, described further below.

a) Show that it is possible for $\mathbb{R}^2-A$ and $\mathbb{R}^2-B$ to be homeomorphic even though $A$ and $B$ are non-homeomorphic closed subsets.

b) If $A \subset \mathbb{R}^2$ is closed and totally disconnected, then $E(\mathbb{R}^2-A)$ is homeomorphic to $A$. Hence if $A$ and $B$ are non-homeomorphic totally disconnected closed subsets, $\mathbb{R}^2-A$ and $\mathbb{R}^2-B$ are non-homeomorphic.

c) The derived set $A'$ of a set $A$ is defined to be the set of non-isolated points of $A$. Show that for each $n$, there is a subset $A_n$ of $\mathbb{R}$ such that the $n$'th derived set ${A_n}^{(n)}$ of $A_n$ consists of a single point.

d) There are $c$ non-homeomorphic closed, totally disconnected subsets of $\mathbb{R}^2$.(Hint: Let $C$ be the cantor set, and $c_1<c_2<c_3\ldots$ a sequence of points in $C$. For each sequence $n_1<n_2<n_3\ldots$, one can add a set $A_{n_i}$ such that its $n_i$'th derived set is $\{c_i\}$.)

e) There are $c$ non-homeomorphic connected open subsets of $\mathbb{R}^2$.

What I've got so far: For part a, $A=$point, and $B=$closed disk should solve the problem. For part c, I think that we can take the $1/n$ sequence (and 0) and add smaller such sequences that converge to each of the points of the original. This can be done recursively.

For part b, I have a feeling that the statement of the problem should be to prove that $E(\mathbb{R}^2-A)$ is homeomorphic to the one-point compactification of $A$ (call it $\tilde{A}$). (The reason I think this is because $A$ might not be compact, but $E(\mathbb{R}^2-A)$ always is. Using the one-point compactification should also take care of the unbounded end.)

I think the idea is to define a function $\tilde{A}\to E(\mathbb{R}^2-A)$ that takes a point $a$ to an end $e$ defined by $e(C)$= the component of $\mathbb{R}^2-A-C$ whose closure in $\mathbb{R}^2$ contains $a$, and takes $\infty$ to the end $e$ defined by $e(C)=$the unique unbounded component of $\mathbb{R}^2-A-C$ (I'm not sure why there's a unique one, but it feels like removing a closed totally disconnected set from a connected open set should leave a connected open set). But I couldn't prove that this function is well-defined, much less bijective, continuous, or open.

If my guess about the one-point compactification in right, this may mess up the "hence" part of b, since I remember reading somehere that it's possible for two non-homeomorphic spaces to have isomorphic one-point compactifications.

For part d, I believe we can take the cantor set, and add the sets in part c vertically over the $c_i$. But I couldn't prove that the result is totally disconnected or closed, or that there are $c$ non-homeomorphic ones.

For part e, I believe that it's enough to prove that the complements of the sets used in part d are connected.

Edit: Beni Bogosel has provided a nice answer to part c below.

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For part $(c)$ you can consider the set $A_n=\{(\sum_{k=1}^n \frac{1}{p_k},0) : p_k \in \{1,2,3,...\}\}$. At each derivation the sum has fewer terms by $1$.

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Thanks, that is very slick. – alephzero314 May 28 '11 at 5:54

There are several references relevant to part (e) at http://mathoverflow.net/questions/25009/counting-submanifolds-of-the-plane

But perhaps there are more elementary solutions. I'd be interested to see them!

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There's a problem with (b), because it might happen that $A$ is compact and totally disconnected and has no isolated points, e.g. the Cantor set; it would not help, as you suggest, to take the "one point compactification" of $A$, since $A$ is already compact. But $E(\mathbb{R}^2-A)$ will always have an isolated point when $A$ is compact, corresponding to the unique end $\infty$ of $R^2$ itself, the "point at infinity". The correct statement of (b) might be just to consider $E(\mathbb{R}^2 - (A \cup \{\infty\})$ minus the point at infinity.

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