Exercise: Let $X$ be the unit circle in $\mathbb{R}^2$; that is $X=\{(x,y):x^2+y^2=1\}$ and has the subspace topology.
Show that $X\setminus\{\langle 1, 0\rangle\}$ is homeomorphic to the open interval $(0,1)$.
Attempt solution:
The exercise claims that $X=\{(x,y):x^2 + y^2=1\}$is a circle but I guess it should be instead $X=\{(x,y):x^2+y^2\leqslant 1\}$.
If it was not a circle then $f:(0,1)\to X\{\langle 1, 0\rangle\}$ would be the same as $f:(0,1)\to X$.
When I saw the circle I thought of using polar coordiantes but it only gets me the circumference $(\cos(\theta),\sin(\theta))$.
I do not know how to address the circle.
Question:
What kind of bijective function could I use on this case?
Thanks in advance!
Best Answer
$f(t) = (\cos 2\pi t, \sin 2\pi t)$ is a possible homeomorphism between $(0,1)$ and $X \setminus \{\langle 1,0\rangle\}$.