Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question

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 $1$.

share|cite|improve this answer
Thanks, that is very slick. – alephzero314 May 28 '11 at 5:54

There are several references relevant to part (e) at

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

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.