Prove that the unit circle is path-connected? I need to show that the unit circle is path connected and connected.  I was able to show that it is connected, by $f:[0,2\pi] \to \mathbb{R}^2$, $f(x)=(r\cos x,r \sin x)$ which is a continuous function. The interval $[0,2\pi]$ is connected and the continuous image of a connected set is connected.  Thus the unit circle is connected.  I'm not sure how to prove path-connectedness, which is the stronger condition.  Thanks!
 A: Hummm, I'm not a native English speaker, so I am a bit lost concerning the mathematical vocabulary, but I'll try my best.
For proving that the unit circle is connected, you could also say that "the only subsets of the unit circle which are both open and closed are the full circle and the empty set". Which is obvious when you think about it.
For the path-connection, I'm taking the definition : a set is path connnected if, for any pair of points you can make a path (continuous function on$[0,1]$) that joins them.
Basically, for the two points $a_1$ and $b_1$ of your comment,
$$f(t)=(r\cos(t*(x_2-x_1)+x_1),r\sin(t*(x_2-x_1)+x_1))$$
is a path that links these two points where we have $f(0) = a_1 \text{ and } f(1)=b_1$.
P.S : This is my first post, so where can I find a faq about how to write mathematical expressions correctly?
A: Your function $f$ can be used to show that the unit circle is in fact path-connected: pick $\mathbf{x}\neq\mathbf{y}$ in the unit circle. Since your $f$ is onto, there exist $a,b\in [0,2\pi]$ such that $f(a)=\mathbf{x}$, $f(b)=\mathbf{y}$. Assume, without loss of generality, that $a\lt b$. Then consider a map $g\colon[0,1]\to[a,b]$ given by $g(t) = a+t(b-a)$. Then consider the function $f\circ g\colon [0,1]\to \mathbf{C}$. 
A: Generally, unit sphere $S^{n-1}$ in $\mathbb{R}^{n}$ defined by the equation $S^{n-1} = \{ \mathbf{x} \mid ||\mathbf{x}|| = 1 \}$ is path connected when $n > 1$. The unit circle in your problem is the special case when $n = 2$.
The justification is given in Munkres's Topology in two examples (Examples 4 and 5 of section 24 "Connected Subspaces of the Real Line", 2nd edition). 
First, to show that the punctured euclidean space defined as $\mathbb{R}^{n} \setminus \{ 0 \}$ is path connected when $n > 1$: Given any two points $\mathbf{x}$ and $\mathbf{y}$ different from $0$, we can join them by the straight-line path $f: [0,1] \to \mathbb{R}^{n} \setminus \{ 0 \}$ defined by $f(t) = (1-t) \mathbf{x} + t \mathbf{y}$ if the path does not go through the origin. Otherwise, the two points can be joined by two paths through a third point.
Then, to show that the unit sphere $S^{n-1}$ is path connected when $n > 1$. Consider the map $g: \mathbb{R}^{n} \setminus \{ 0 \} \to S^{n-1}$ defined by $g(\mathbf{x}) = \mathbf{x} / ||\mathbf{x}||$. $g$ is continuous and surjective. The conclusion follows from the fact that the continuous image of a path connected space is path connected.
A: This is a proof that does not use any trigonometry. The technique can be easily extended to higher dimensional spheres.
If $(x_0,y_0)$ belongs to the unit circle, the mapping
$\tag 1 t \mapsto \left(\,(1-t)x_0,\, \pm\sqrt{1 - [(1-t)x_0]^2}\,\right)$
defines a path connecting $(x_0, y_o)$ to either $(0, 1)$ or $(0, -1)$.
We can path-connect  $(0, -1)$ to $(0, 1)$ via the mapping
$\tag 2 t \mapsto \left(\,\sqrt{1-t^2},\, t\,\right)$
defined on $[-1,+1]$.
So a path can be found that connects any point $(x_0,y_0)$ on the unit circle to the point $(0,1)$. Since we can traverse any path in the opposite direction, a path can be found to connect any two arbitrary points.
