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?



Can the nth projective space be covered by n charts?
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:


1. 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 $\hat{X}$.
2. 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 $H^n({\mathbb RP}^{n-1},{\mathbb Z}/2{\mathbb Z})=0$.


