Disprove Homeomorphism 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.
 A: 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.
A: 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.
A: 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 $B$.
A: 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 $ ?
