Proving $R^2$ \ ${(0,0)}$ is pathwise connected and connected 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.
 A: 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.
A: 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.
A: There are elementary proofs (no trigonometry required) that the circle is path-connected.
Any point $(x_0,y_0)$ in the punctured plane belongs to a circle $x^2 + y^2 = r^2$. So it can be path connected to $(0,r)$. And certainly there is a path connecting $(0,r)$ to $(0,1)$. Since we can traverse any path in the opposite direction, any two points in the punctured plane can be connected with a path.
