# 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!

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You have to show that any two points on the unit circle can be joined by a path (itself on the unit circle) that connects them. Perhaps you can use a variant of your function $f$ to show this. –  Jesse Madnick Apr 6 '11 at 15:23
Also, there's no need for the $r$ in your function $f$. (That is, one should have $r = 1$.) –  Jesse Madnick Apr 6 '11 at 15:34
Yes you're correct. I was using the general form. Let a1=(cosx1,sinx1) and a2=(cosx2,sinx2) I want to show that there is a path connecting these two points. I'm still struggling to develop this path –  hawaii99 Apr 6 '11 at 15:42
Dear Caligirl, Just to expand slightly on Jesse Madnick's comment: in your question you wrote down a path that covers the whole unit circle! Just restricting the domain (i.e. replacing $[0,2\pi]$ by an appropriately chosen subinterval) you can find a path that joins any two points. Regards, –  Matt E Apr 6 '11 at 16:09
You could use the same argument: the continuous image of a path connected space is path connected. –  Robert Bell Apr 6 '11 at 19:23

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 $f\circ g\colon [0,1]\to \mathbf{C}$.

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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, $f(x)=(r\cos(t*(x_2-x_1)+x_1),r\sin(t*(x_2-x_1)+x_1))$ is a path that links these two points ($f(0) = a_1 \text{ and } f(1)=b_1$).

P.S : This is my first post, so where can I find a faq about how to write mathematical expressions correctly?

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For some suggestions on equations, you can see meta.math.stackexchange.com/questions/1773/… –  Ross Millikan Apr 6 '11 at 16:00
I formatted your answer, please check that it still says what you meant. To see what I did, you can right-click on any piece and choose Show Source. There are single dollar signs around that to tell the page to render it. Double dollar signs gives display mode. Welcome aboard. –  Ross Millikan Apr 6 '11 at 16:05
Thanks to both of you! I guess my Latex reports will be of some use from now on! –  Fezvez Apr 6 '11 at 16:09
Both of me? .... :) –  Ross Millikan Apr 6 '11 at 16:13
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 $n = 2$.
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.