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.
Best Answer
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.