A variety $X$ by definition has an open cover with finitely many affine varieties $X_1, \dots, X_n$. The claim is that dimension of $X$ is max of dim $X_i$.
I am not getting anywhere with this problem. So I start with a chain of irreducible closed subsets $\{F_k\}_k$ in $X$. Their intersections $\{F_k \cap X_i\}_k$ will also be closed in the subspace topology of $X_i$, and they would also be irreducible since any open set $F_k \cap X_i$ of an irreducible set $F_k$is (is it true?). But there is no guarantee on strictness of the chain. Conversely, if i start with a chain of irreducible closed sets in one of the $X_i$, I can't guarantee that I can get a chain of irreducibles in $X$. Any hint would be appreciated. Thanks.
What is your definition for the dimension of a variety? what about the dimension of an affine variety? An approach is the following
$dim X=max_x dim_xX$ (where $dim_xX$ is the dimension of $X$ at $x$)
Moreover for affine varieties, $dimX_i=dim_xX_i$ for all $x\in X_i$ and therefore the result follows.