4
$\begingroup$

I am studying Real Analysis with the book "Real mathematical analysis" and it has been a very challenging (but not impossible) read for me so far. However, some of the exercises are pretty brutal, so one I have been stuck on is (page.129/exercise 47)

Suppose $A,B\subseteq\mathbb{R}^2$
(a) If $A$ and $B$ are homeomorphic, are their complements homeomorphic
(b) If $A$ and $B$ are homeomorphic and compact, are their complements homeomorphic
(c) If $A$ and $B$ are homeomorphic, compact and connected, are their complements homeomorphic

As the conditions get progressively more restrictive, I am very sure, that only the last one has a chance of being true and indeed, it is. For the first one, I simply used the fact, that $(0,1)$ is homeomorphic to $\mathbb{R}$, but their complements are not homeomorphic, and set $A=(0,1)\times \{0\}$ and $B=\mathbb{R}\times\{0\}$, where the complement of $A$ is connected, while the complement of $B$ is not. However, with the added restriction of compactness, I can't disconnect $\mathbb{R}^2$, as that would require an unbounded set, and after 2 hours of banging my head against the wall, I still can't find a counterexample and so I want to ask here for one. I can probably wager, that both $A$ and $B$ are disconnected, and that the counterexample will arise from the complement of $A$ being disconnected, while the complement of $B$ will be connected, as that has been the only tool mentioned so far for distinguishing non homeomorphic sets.

$\endgroup$
3
  • $\begingroup$ What book is that? You must mention the author, at least, when citing a book or any publication. (Good question in any case). $\endgroup$
    – Giuseppe Negro
    Commented Sep 17, 2023 at 22:36
  • $\begingroup$ @GiuseppeNegro Its "Real Mathematical Analysis" 2nd edition by Charles C. Pugh, published by Springer, I thought the info was sufficient, but here are the details in any case $\endgroup$
    – Brummi
    Commented Sep 18, 2023 at 5:54
  • $\begingroup$ I digress a bit on this point. I would say, at least cite the author. "Pugh's textbook in analysis" is much more informative than "Real Mathematical Analysis". The same goes with any publication. People tend to remember the author first and other info later. Unlike in literature, in scientific publications the title is actually not often remembered. $\endgroup$
    – Giuseppe Negro
    Commented Sep 20, 2023 at 9:16

1 Answer 1

Reset to default
5
$\begingroup$

Let $S$ be the unit circle, let $X=S\cup \{(0,0)\}$, $Y=S\cup \{(0,2)\}$.

Then $X$ and $Y$ are certainly homeomorphic, but their complements are not, as one has a simply connected component and the other does not.

Update

For a simple proof that a punctured disk is not homeomorphic to an unpunctured disk, which avoids algebraic topology, we can argue as follows.

First, since the disk is homeomorphic to the plane, its enough to show the punctured plane is not homeomorphic to the plane itself.

To this end, observe that every compact subset of the plane has exactly one complementary component that is not relatively compact - this follows from the Heine-Borel theorem that compactness in $\mathbb R^n$ is equivalent to being closed and bounded.

On the other hand, neither of the two complementary components of the unit circle are relatively compact in the punctured plane.

Since the property "Every compact subset has exactly one complementary component that is not relatively compact" is clearly preserved by homeomorphisms (it is stated entirely in terms of topological properties), this proves the two spaces are not homeomorphic.

Remark

The aforementioned proof does not actually prove the original statement, as FShrike points out in the comments. One final step is necessary - show that each of the two complementary components of $X=S\cup \{(0,0)\}$ is homeomorphic to a punctured disk. To do this is not hard - one component literally is the punctured disk, and the other can be mapped to the punctured plane via the map $re^{i\theta}\mapsto (r-1)e^{i\theta}$, and from there to the punctured disk in the usual fashion.

$\endgroup$
8
  • $\begingroup$ Thanks, I did think of that, but I (preemtively) dismissed it, as their complements are both connected, and I couldn't prove that a homeomorphism does not exist, although it seemed like it. Do you have a link to a paper/article that lays out some basic criterea (you used the term "simply connected", which I haven't heard yet) for telling, that sets are not homeomorphic? I'm reading up on it, and am planning to post a detailed proof of the counterexample for future users, that are also struggling on the exercise. Thanks for your answer :) $\endgroup$
    – Brummi
    Commented Sep 17, 2023 at 8:45
  • 1
    $\begingroup$ @Brummi: there are essentially no tools available in an introductory real analysis course for showing that two topological spaces are not homeomorphic, aside from maybe simple arguments involving connectivity and neighborhoods, none of which are available here. Simple connectivity refers to the fundamental group (en.wikipedia.org/wiki/Fundamental_group) which is an algebraic topology subject; I wouldn't worry about it for now. $\endgroup$
    – Qiaochu Yuan
    Commented Sep 17, 2023 at 8:48
  • $\begingroup$ @QiaochuYuan I thought of one way to avoid the more complex machinery, posted it in the update. $\endgroup$
    – M W
    Commented Sep 17, 2023 at 9:38
  • $\begingroup$ More work would be needed as you have to show the union of a disk with a punctured unbounded annulus $\Bbb R^2\setminus D^2$ is not homeomorphic to the union of a punctured disk with an unpunctured annulus. Maybe the hypothetical homeomorphism doesn’t take one disk to the other. N.B. this isn’t too hard to rule out but it needs to be done $\endgroup$
    – FShrike
    Commented Sep 17, 2023 at 9:40
  • $\begingroup$ @Brummi You don’t need to dig into much algebraic topology at all to learn about distinguishing a circle from a disk. The calculation of $\pi_1 S^1$ can be done without too much bother, one only needs a general-topological lemma about lifting to covering spaces. $\endgroup$
    – FShrike
    Commented Sep 17, 2023 at 9:44

You must log in to answer this question.

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