I'm currently working on a topology assignment which is, unfortunately, due today.
As part of that, I need to show that one-point compactification of the real line, $\mathbb{R}\cup\infty$ is homeomorphic to the unit circle.
I've come so far to have defined a function $f$ with
$(x,y)=(\cos(2\arctan(t)),\sin(2\arctan(t)))$, which should map the reals onto the unit circle and be bijective and continuous. Now since my domain is compact, if I can show that the unit circle is Hausdorff, I can conclude that my function is a homeomorphism, correct?
However, which topology would I use for that?
Also, I am having a hard time trying to prove that f is surjective. Thought about dividing it up into two functions, from reals to $(-\pi,\pi)$ and then to the unit circle, but that doesn't really seem to work either.
Any thoughts?
Topology is confusing…
(Intuitively, I absolutely see why all this should be the case, however I am struggling with the formal stuff.)
Edit:
Okay I have tried the approach of defining an inverse function, suggested and then proving that it is continuous.
So far I've got:
$$g=\tan\left(\frac{1}{2}\arctan\left(\frac{y}{x}\right)\right)$$
Problem is, that this function is not bijective, so there is something missing (case distinction?) but I can't figure out what…
At least within $\left(\frac{-\pi}{2},\frac{\pi}{2}\right)$ it seems to work, and it's easy to show that combining f and g gives the identity.
Any tips maybe?
Cheers
Tom
Best Answer
I don't know why I didn't say the following at the outset. I guess I was just going along with your method.
Rid your proof of trigonometric functions and instead do the following:
$$ t\mapsto (x,y) = \begin{cases} \phantom{\lim\limits_{t\to\infty}} \left( \dfrac{1-t^2}{1+t^2}, \dfrac{2t}{1+t^2} \right) & \text{if }t\ne\infty, \\[10pt] \lim\limits_{t\to\infty} \left( \dfrac{1-t^2}{1+t^2}, \dfrac{2t}{1+t^2} \right) & \text{if }t=\infty. \end{cases} $$
That's a homeomorphism from $\mathbb R\cup\{\infty\}$ to $\{(x,y)\in\mathbb R^2:x^2+y^2=1\}$.
To show that it's surjective, show that $t=\dfrac{y}{x+1}$ (and notice that $\dfrac{y}{x+1}\to\infty$ as $(x,y)\to(-1,0)$ along the curve $x^2+y^2=1$).