Prove that for $k$ a finite field every subset is closed (and open) in Zariski topology

Question

The affine algebraic sets define a topology on the affine $n$-space, which is called the Zariski topology. I think that the algebraic sets are the closed sets in the Zariski topology, so I am not very sure of what I have to prove. Can someone please help me ? Also what if $k$ is not finite then what will happen?

Answer 1

Well, every point $(a_1,\dotsc,a_n)$ is closed since it is the zero set of the polynomials $x_i-a_i$ for $1\leq i \leq n$. Now just use that finite unions of closed sets are closed.

This turns out to be wrong already for $\mathbb{A}^1_k$ if your field is infinite as the closed sets are exactly given by $V(0) = \mathbb{A}^1_k$ and the finite subsets of it (every non-zero polynomial in one variable has finitely many roots).

Answer 2

The set $\{-1,1,42\}\subset \Bbb R$ (or $\Bbb C$ or $\Bbb Q$, ...) is Zariski closed because it is the set of roots of $X^3-42X^2-X+42$. Similarly, every finite set of a field (i.e., of affine $1$-space) is closed. For a finite field, that's all subsets. And their complemets are the open sets ...