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I need some help (hints or an answer) in finding the actual equation of this bijections from the reals $ \mathbb{R} $ to $ (0,1) $. We may assume that the radius of the circle is $ \dfrac{1}{2} $.

enter image description here

Also, is it possible to transform this geometric representation to an order-preserving isomorphism?


Also, I am aware there are much simpler isomorphisms between $ \mathbb{R} $ and $ (0,1) $.

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See Stereographic projection at for inspiration. – Babak S. Jan 23 '13 at 6:59
"(This is my first post!)" Then what's this? – Rahul Jan 23 '13 at 7:45
up vote 2 down vote accepted

The point $x$ in $(0,1)$ is sent to the point $y(x)$ in $\mathbb R$ if there exists some angle $\theta$ in $(0,\pi)$ such that $\cos\theta=2x-1$ and $y=a\cot\theta$, where $a$ is the distance between the center of the circle and the line. Since $\sin\theta=\sqrt{1-\cos^2\theta}$, this yields $$ y(x)=\frac{a(2x-1)}{\sqrt{1-(2x-1)^2}}=\frac{a(2x-1)}{\sqrt{4x(1-x)}}. $$ For $y\mapsto x(y)$ sending $\mathbb R$ to $(0,1)$, use the inverse mapping $$ x(y)=\frac12\left(1+\frac{y}{\sqrt{y^2+a^2}}\right). $$ Both functions $x\mapsto y(x)$ and $y\mapsto x(y)$ are increasing diffeomorphisms.

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Thank you sir! great answer! – Cornelius Johnson Jan 23 '13 at 8:46

Just do the work.

Assume that the circle is $x^2+y^2 = 1$. Assume that the line is $y=-k$ for some positive real number $k$.

Then, the point $(x, k)$ will get mapped to the point $(\frac {x}{\sqrt{x^2+k^2}}, \frac {-k} { \sqrt{x^2+k^2}})$.

If you want to map it to $(0, 1)$ as opposed to $(-1, 1)$, then take the transformation $t \rightarrow \frac {t+1} {2} $. Alternatively, use another circle but that makes the derivations ugly.

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Calvin I dont see how this represents a bijection from R to (0,1). Since (x,k) doesn't lie in R. I am probably just not seeing it, could you define an explicit function? – Cornelius Johnson Jan 23 '13 at 7:15
the map $f(x) = \frac {x}{\sqrt{x^2+k^2}}$ is a map from $\mathbb{R}$ to $(-1, 1)$. That was implicit in your diagram. – Calvin Lin Jan 23 '13 at 7:17
Oh ok thanks you so much. You are right. Sorry I forgot to mention the circle has radius 1/2, which implies the top line segment is actually (0,1) not (-1,1) – Cornelius Johnson Jan 23 '13 at 7:23
@CorneliusJohnson It's just much easier to do it on $(-1, 1)$, since that's the standard unit circle that we're used to. As I said, if you want to use $(0, 1)$, just do a further transformation. The direct approach will give you quite ugly algebra that takes away the intuition of what's happening. – Calvin Lin Jan 23 '13 at 7:30

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