# Why can a projective variety of dimension $n$ be covered by $n+1$ affine open subsets?

My question: Why can a projective variety of dimension $$n$$ be covered by $$n+1$$ affine open subsets?

I can see the result holds when the variety is $$\mathbb P^n$$ or a hypersurface $$X$$ in $$\mathbb P^{n+1}$$. The second is true because you can suppose the point $$[0:\cdots:0:1] \not \in X$$ and then cover $$X$$ with $$A_0, \ldots,A_n$$, where $$A_i = \{[x_0: \cdots:x_{n+1}]: x_i \neq 0\}.$$

Obs.: The reason why I want to prove this fact is to prove that the Čech cohomology $$H^p(X,F)$$ of a projective variety $$X$$ of dimension $$n$$ is $$0$$ for $$p>n$$, where $$F$$ is a quasicoherent sheaf.

Let $$X_d\subseteq \mathbb{P}^n$$ be a projective variety of dimension $$d$$. Pick a hyperplane $$H_d$$ which does not contain any irreducible component of $$X_d$$. Then $$X_{d-1}=X_d\cap H_d$$ has dimension $$d-1$$. We can then repeat this process and pick a hyperplane $$H_{d-1}$$ which does not contain any irreducible component of $$X_{d-1}$$, and define $$X_{d-2}=X_{d-1}\cap H_{d-1}$$. We repeat this until we reach $$X_0$$ which is $$0$$-dimensional and finally have $$H_0$$ which is disjoint from $$X_0$$. We then see that every point of $$X_d$$ is not in $$H_i$$ for some $$i=0,\dots,d$$ so $$X_d$$ is covered by the $$d+1$$ affine open subvarieties $$X_d\setminus H_i$$.