Homeomorphism from real interval to an arc of a circle I haven't seen this question anywhere, surprisingly.
In a proof of some theorem, my lecture note abruptly states the above. That

Since there is a homeomorphism of any real interval and the arc of a unit circle with points $a,b$ as end points...

I tried. I really did. But I can't find one.
Issue is, I don't know how to think of it. The real interval's straight forward, sure just say $[s_0,s_1] \in \mathbb{R}$(though, the term "interval" is very obscure, does it mean open?closed? regardless?)
But the unit circle, I can't make up my mind as to what I should work with. As a circle in $\mathbb{C}$ so I express any point on the circle by $\cos{\theta}+i \sin{\theta}$? How could I determine the points $a,b$, should I restrict angle $\theta$ as I dunno, $\theta_0 \leq \theta \leq \theta_1$? Or similarly, should I use polar coordinates $(\cos{\theta},\sin{\theta})$? Ot should I use vectors? Say a point $p$ on the unit circle as $p=(x,y)^T$? What's best here?
If I use $p=\cos{\theta}+i \sin{\theta}$ then, I need to make some $s\in[s_0,s_1]$ correspond with the "turn" of the points along the unit circle continuously, so perhaps it'll be in some form $(\cos{\theta}+i \sin{\theta})^s$ or something by De Moivre? But I have to make sure it will not go past the end points $a,b$ so...If I let $\theta_1=k\theta_0$ for some $k \in \mathbb{R}$, then I need to make sure for any $s$ that $\theta$ doesn't exceed $\theta_1$ so I need to make $s$ related to $k$...
No, this isn't going anywhere. But it's exploding my brain with mines. 
Please someone help me before this homemophism shreds my brain into pieces.
 A: Draw a unit circle and a straight line below it, now choose a point from the circle different than the "north pole". Now from the "north pole" of the circle, draw a straight line to the selected point of the circle and prolong it until it intersects the line you drew below the circle. This map, called the $\textbf{stereographic projection}$, defines an homeomorphism between the real line and the circle minus a point (the "north pole"). Using this, select any arc and the image of the arc under this map is an interval (open or closed depending on if the arc you took is open or closed). Since every open (closed) interval is homeomorphic to any other open (closed) interval, you get the result.
A: This answer assumes that the interval is bounded.
Consider the semicircular arc given by the portion of the curve $x^2+y^2=0$ above or on the $x$-axis. (You can think of this as the set of points $\langle\cos\theta,\sin\theta\rangle$ for $0\le\theta\le\pi$.)
Consider, too, the interval $[-1,1]$ on the $x$-axis.
The projection onto the $x$-axis, mapping $\langle x,y\rangle$ on the circle to $x$ in the interval, or mapping $\langle\cos\theta,\sin\theta\rangle$ to $\cos\theta$, gives a homeomorphism from the semicircle to the interval.
(If you need open intervals and arcs, just ignore the endpoints; the same projection works.)
From here, all you need is a homeomorphism from any real interval to $[-1,1]$ (or $(-1,1)$ if one wants open intervals) and a homeomorphism from the semicircle to any arc.
