Let us consider two subspaces of $(\mathbb R^2, \tau)$: $A=\{(x,y)\in \mathbb R^2: 0<x<1,0<y<1\}$ and $B=\{(x,y)\in \mathbb R^2: x^2+y^2<1\}$. Here $\tau$ denotes the usual topology on $\mathbb R^2$. I want to show that $A$ and $B$ are homeomorphic, but don't know how to define a homeomorphism between these sets. Any hint will be helpful. Thanks!
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4One possible way is to show that both are homeomorphic to $\mathbb{R}^2$. – Daniel Fischer Oct 19 '13 at 17:30
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Another approach is to begin by translating the rectangle to put its centre at the origin. Then expand radially in each direction by a factor that takes the boundary of the square to the unit circle. E.g., along the axes the expansion factor will be $2$; along the diagonals, $\frac1{\sqrt2}$. The expansion factor in the direction $\theta$ varies continuously with $\theta$. This gives you a homeomorphism of $A$ onto $B$.
Brian M. Scott
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@ B.M. Scott: Sir, it would be very much helpfull if you kindly elaborate the explanation a little. – Anupam Oct 20 '13 at 09:32
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In the polar direction $\theta$ the distance $r(\theta)$ from the origin to the boundary of the translated square is $\sec\theta$ if $0\le\theta\le\pi/4$, $\csc\theta$ if $\pi/4\le\theta\le3\pi/4$, etc. Consider the map that sends $\langle r,\theta\rangle$ in the translated square to $\left\langle\frac{r}{r(\theta)},\theta\right\rangle$: it expands the translated square to the open disk. – Brian M. Scott Oct 20 '13 at 09:36
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For the rectangle, try scaling/translating, then applying arctan to the components. For the disk, try $\frac{x}{1-|x|}$. These give homeomorphisms onto $\mathbb{R}^2$.
user91823
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