In his book Fundamentals of Algebraic Topology Weintraub claims on page 96:
For $k=\mathbb R$ or $\mathbb C$, $\mathbb kP^n\setminus [1,0,\cdots,0]$ is homeomorphic to $k^n$.
This look fishy to me, in particular because for $k=\mathbb C, n\geq 2$ the assertion is completely false in the holomorphic category.
However I can neither prove nor disprove Weintraub's statement in the topological category.
So, is the displayed assertion true or false?
As mentioned by Tyrone in his comment (which I upvoted) , Weintraub is indeed completely wrong because $X=k\mathbb P^n\setminus [1,0,\cdots,0]$ is homotopic to $k\mathbb P^{n-1}$, which of course is not homotopic to the contractible space $k^n$ for $n\geq 2$.
A different from Tyrone's way to see the homotopy is by noticing that $k\mathbb P^{n-1}$ (identified to the the hyperplane $x_0=0$ of $k\mathbb P^n$) is a strong deformation retract of $X$ under the homotopy $(t,[x_0,x_1,\cdots,x_n])\mapsto [tx_0,x_1,\cdots,x_n]$.