Can the nth projective space be covered by n charts? 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?
 A: Expanding on the comment by @user127776, the key reference is Palais, "Lusternik-Schnirelman Theory on Banach Manifolds", Topology 5 (1966),
where it is proved that if $X$ can be covered by $n$ contractible closed sets, then the cup-length of $X$ is strictly less than $n$.
(Here the cup-length is the largest $n$ such that for some field $F$ and some elements $c_1,\ldots,c_n$ in $H^*(X,F)$, we have $c_1\cup\ldots\cup c_n\neq 0$.)
This rules out covering ${\mathbb RP}^n$ with $n$ closed contractible sets, which should suffice here (after slightly shrinking the given $n$ copies of ${\mathbb R}^n$).
Editing to add:
More generally, suppose $X$ is a compact Hausdorff space covered by $n$ closed sets $X_1,\ldots, X_n$ with all $H^1(X_i,{\mathbb Z}/2{\mathbb Z})=0 $.  (Equivalently, any (real) line bundle on $X_i$ is trivial.)
Theorem. Any line bundle on $X$ can be generated by $n$ sections.
Proof. Let $\hat{X}= Spec(C(X,{\mathbb R}))$, so that $X$ imbeds in $\hat{X}$.  Note that:

*

*Because $X$ is normal, each $X_i$ is defined by the vanishing of a continuous function, so the $\hat{X}_i$ form a closed covering of $\hat{X}$.


*By Swan's theorem, the map that takes a vector bundle over $\hat{X}$ to its pullback over $X$ is an equivalence of categories (and likewise with $X$ replaced by $X_i$).
Now because every line bundle on $X_i$ is trivial, so is every line bundle on $\hat{X}_i$.
Because $\hat{X}$ is an affine scheme, a line bundle corresponds to a projective module, which in turn is the image of an idempotent matrix with entries in $C(X,{\mathbb R})$.  A little thought reveals that this matrix can be taken to be $n\times n$.  It follows that any line bundle on $\hat{X}$ is generated by $n$ sections.  Therefore (by the Swan correspondence) so is any line bundle on $X$, as advertised.
 Corollary. For any $c\in H^1(X,{\mathbb Z}/2{\mathbb Z})$, the $n$-fold cup product $c^n\in H^n(X,{\mathbb Z}/2{\mathbb Z})$ is zero.
Proof. $c$ is the first Stiefel-Whitney class of some line bundle $\xi$.  Let $\phi_\xi:X\rightarrow {\mathbb RP}^\infty$ be the classifying map of $\xi$.  The $n$ sections guaranteed by the theorem provide a factorization of $\phi_\xi$ through ${\mathbb RP}^{n-1}$.  But $H^n({\mathbb RP}^{n-1},{\mathbb Z}/2{\mathbb Z})=0$.
A: 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.
