I have been reading about invertibility of functions and one subtopic is branches of functions/inverses. While reading about defining branches of the argument, a statement was left as an exercise to prove.
First it says: "A branch of $\arg(x,y)$ can be defined on an open set $T$ of the $xy$-plane if and only if the polar coordinates of the points in $T$ form an open subset of the $r\theta$-plane that does not intersect the $\theta$-axis or contain any two points of the form $(r,\theta)$ and $(r,\theta+2k\pi)$, where $k$ is a nonzero integer. No subset containing the origin has this property..."
Which I understand, but then it says the following: "...nor does any deleted neighborhood of the origin"
Which is left as an exercise to prove. I have no inclination to why this would be true so I dont know where to even start. For all points in a deleted neighborhood of $(0,0)$ all points would be mapped to a set with no points $(r,\theta)$ and $(r,\theta+2k\pi)$ and all point have non-zero modulus so I see no contradiction there.