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I can see that it is clear. Between any two points we can find a path that does not cross origin, so it is pathwise connected, then it is connected.

But I am having problems with the proof. Can you give me some hints?

Thank you.

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You say that for any two points, it's clear you can find a path between them. So just write down those paths. – Chris Eagle Dec 9 '11 at 20:57
It is easier to find a path between the two points if you write them in polar coordinates. – Thomas Andrews Dec 9 '11 at 21:01
(I mean, easier to write down a formula for the path.) – Thomas Andrews Dec 9 '11 at 21:05
up vote 2 down vote accepted

Let $a_0,a_1\in\mathbb R^2\setminus 0$. Then write $a_i=r_i(\cos \theta_i,\sin \theta_i)$ in polar coordinates, with $r_i>0$. Define $f(t) = r(t)(\cos \theta(t),\sin \theta(t))$ where $r(t)=(1-t)r_0 + tr_1$ and $\theta(t)=(1-t)\theta_0 + t\theta_1$.

Then $\theta(0)=\theta_0$ and $\theta(1)=\theta_1$, $r(0)=r_0$ and $r(1)=r_1$, so $f(0)=a_0$ and $f(1)=a_1$. Also, for $t\in(0,1)$, $r(t)>0$ since $t,1-t,r_0,r_1>0$. So $f(t)\neq 0$ for all $t$, and therefore $f$ defines a path from $a_0$ to $a_1$.

Basically, $\mathbb R^2\setminus 0$ is the continuous image of the set $\mathbb R^+\times \mathbb R$, which is convex and hence path-connected. The continuous image of a path-connected space is a path-connected space.

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Here is a solution sketch that should help.

Fix two arbitrary points $a,b$ in $\mathbb{R}^2$/${(0,0)}$. If the line between them does not pass through the origin, then we are done. Simply parameterize the line between them.

Otherwise, we know $a,0,b$ are collinear. Construct a circle around the origin small enough so that it contains the origin but does not contain $a$ or $b$. Construct the path between $a$ and $b$ in three parts.

1) Move along the line between $a$ and $b$ until you intersect the circle.

2) Continue along the circle $\pi$ radians until you are diametrically opposite from where you first intersected the semicircle.

3) Continue along the straight line path to $b$.

This path should be easy to construct using the standard parametric formula for a circle combined with the line equation you found above. Now prove it doesn't intersect $(0,0)$ and you're done. As you note, once you prove it is path-connected, it follows that the space is connected.

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Or just pick a point $c\ne 0$ on the normal to $\overline{ab}$ at $0$ and construct the segments $\overline{ac}$ and $\overline{cb}$. But @Thomas Andrews offers an even easier approach. WLOG $|a|\le|b|$, so we move radially out from $a$ to $c$ s.t. $|c|=|b|$, then in a circular arc to $b$. – Brian M. Scott Dec 9 '11 at 21:19
@BrianM.Scott Or, as in my answer below, you can actually write out a formula which does both simultaneously. Basically, this amounts to the fact that the continuous image of a path-connected space is necessarily path-connected. – Thomas Andrews Dec 9 '11 at 21:34

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