TITLE: Can the nth projective space be covered by n charts?
QUESTION [38 upvotes]: That is, is there an open cover of $\mathbb{R}P^n$ by $n$ sets homeomorphic to $\mathbb{R}^n$?
I came up with this question a few years ago and I´ve thought about it from time to time, but I haven´t been able to solve it. I suspect the answer is negative but I´m not very sure. Also, is there an area of topology which studies questions like this one?

REPLY [16 votes]: It seems worth giving the cup-length argument, as it's relatively short and sweet.
Suppose $\mathbb{R}P^n=U_1\cup\cdots\cup U_n$, with each $U_i\approx\mathbb{R}^n$, and let $c\in H^1(\mathbb{R}P^n;\mathbb{Z}/2)$ be the generator.
For each $i$ the inclusion-induced map $H^1(\mathbb{R}P^n;\mathbb{Z}/2)\to H^1(U_i;\mathbb{Z}/2)$ is trivial, so by the long exact cohomology sequence of the pair $(\mathbb{R}P^n,U_i)$ there exists a relative cohomology class $c_i\in H^1(\mathbb{R}P^n,U_i;\mathbb{Z}/2)$ whose image in absolute cohomology is $c$. But then by the naturality of relative cup products, $c^n$ is the image of
$$
c_1\cdots c_n\in H^n(\mathbb{R}P^n,U_1\cup\cdots\cup U_n;\mathbb{Z}/2)=0,
$$
and therefore $c^n=0$, a contradiction.
As Aleksander Milivojevic points out in the comments, the relevant area of topology is the study of Lusternik--Schnirelmann category and related invariants.