I have a problem that puzzles me. I need to show that the two sets $A = \{(x,y) \in \mathbb{R}^2 \, \, \vert \, \, |x| \leq 1 \}$ and $B = \{(x,y) \in \mathbb{R}^2 \, \, \vert \, \, x \geq 0 \}$ are not homeomorphic; but I'm not able to figure out how start or what I need to arrive at.
Disprove Homeomorphism
4 Answers
The one-point compactification of $A$ is homeomorphic to the space $X$ obtained by identifying the points $\langle 0,1\rangle$ and $\langle 0,-1\rangle$ of the disk $D=\{\langle x,y\rangle:|x|+|y|\le 1\}$. The one-point compactification of $B$ is homeomorphic to $D$ itself. $X$ and $D$ are not homeomorphic, so $A$ and $B$ cannot be homeomorphic. However, I don’t immediately see a way to prove that $X$ and $D$ are not homeomorphic without using homotopy.
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0why don't we just say that $\partial A$ is connected and $\partial B$ is not? – 2012-04-07
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1@Norbert: Consider $A$ and $B$ as spaces independent of their embeddings in the plane. How do you define their boundaries? – 2012-04-07
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0@Brian: You probably mean $"\le"$ instead of $"="$ in the equation of $D.$ – 2012-04-07
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0BrianM.Scott and @Norbert But I don't -- why are we not allowed to embed $A$ and $B$ into $\mathbb R^2$? – 2012-04-07
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0@Xabier: I sure do; thanks. – 2012-04-07
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1@Matt: Simpler example: $(0,1)$ can be embedded in $\Bbb R$ as itself or as $(0,\to)$, and these sets have different boundaries in $\Bbb R$, $\{0,1\}$ versus $\{0\}$. Here the different boundaries of the embedded sets don’t imply that the embedded sets aren’t homeomorphic. – 2012-04-07
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0@BrianM.Scott Thank you! : ) – 2012-04-07
Look at the following three disjoint sets that cover $A$: $S_1 = \{(x,y), y>0\}\subset A, S_2 = \{(x,y), y=0\}\subset A, S_3 = \{(x,y), y<0\}\subset A$. Now $A-S_2$ is disconnected and a union of $S_1$ and $S_3$ whereas the image of $S_2$ has to be compact. Use this to conclude the proof.
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3I don’t see how this is supposed to work. $S_1$ and $S_3$ are homeomorphic to $A$. $S_2$ is homeomorphic to $$T=\{\langle x,y\rangle\in B:x^2+y^2=1\}\;,$$ whose removel splits $B$ into two pieces each homeomorphic to $B$. In order to show that the homeomorphism of $S_2$ to $T$ can’t be extended to $A$, it seems to me that you already need to know that $A$ is not homeomorphic to $B$. – 2012-04-07
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1@Brian, $S_1$ and $S_3$ do not have compact closure in $A$. That is, $A$ can be separated into two non-precompact connected components upon removal of a compact set; $B$ cannot be. ($A$ has two ends, $B$ has one.) – 2012-04-07
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0@ccc: That answers my specific objection, but it seems to me that M.B.’s argument is still incomplete, even after your comment is included. Specifically, how does one prove the impossibility of a similar decomposition of $B$? I may be missing something simple, but I don’t see it. – 2012-04-07
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0@Brian. Doesn't this follow from the fact that a subset of the plane is compact if and only if it is closed and bounded? Since a compact set lives in a circle of radius $r$ about the origin, there can be only one connected component containing points outside that circle. That means the other connected component must be bounded, thus precompact. – 2012-04-07
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0@ccc: Fair enough: I was missing something fairly simple. – 2012-04-07
Call a point $p\in A\ $ special $\ $ if $A\setminus\{p\}$ is simply connected, and let $\partial A$ be the set of these special points; similarly for $B$. A homeomorphism $\phi:\ A\to B$ would have to map $\partial A$ onto $\partial B$.
The special points of $A$ are the points on the lines $|x|=1$, and the special points of $B$ are the points on the line $x=0$. Now there is a line $\ell\subset A\setminus\partial A$ that separates some special points of $A$ from others, but there is no such curve in $B$.
What about this: if A and B are not homeomorphic, then there exists a continuous bijection $ f: \, A \rightarrow B $ such that $ f^{-1} $ is not continuous. Then look at the function $f(x,y) = \left( \tan\left( \frac{(1+x) \pi}{4} \right), y \right)$. Problem at $ x = 1 $ ?
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0That's not enough: there may be plenty of continuous bijections with discontinuous inverse. It suffices to have one continuous bijection with continuous inverse. – 2012-04-07