Building a circle homeomorphism with $(0,1)$

Question

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!

Answer 1

$f(t) = (\cos 2\pi t, \sin 2\pi t)$ is a possible homeomorphism between $(0,1)$ and $X \setminus \{\langle 1,0\rangle\}$.

Answer 2

It cannot be that $x^2+y^2\le 1$ be homeomorphic to a line since a plane (or a part of a plane) cannot be homeomorphic to a line (check out their Lebesgue measure!) Also a homeomorphism preserves continuity of mapping. A possible such mapping is $f(t)=(\cos 2\pi t,\sin 2\pi t)$ thanks to @Henno Brandsma (think why and how such a mapping is continuous!)