Affine space over algebraically closed field cannot be written as a finite union of lines. Let $k$ be an algebraically closed field and $\mathbb{A}^n=\{(a_1,\dots,a_n):a_i\in k\}$ be an affine space. I want to show that $\mathbb{A}^n$ cannot be written as a finite union of lines where $k$ is algebraically closed for $n>1$.
By contradiction: let $\mathbb{A}^n=l_1\cup\dots\cup l_n$ where $l_i$ is the line in $\mathbb{A}^n$ ($l_i$ is a closed subset of $\mathbb{A}^n$). Then we get a contradiction since $\mathbb{A}^n$ is an irreducible topological space, so it cannot be written as a finite union of proper non-empty closed subsets.
However, I am confused if I need $k$ to be algebraically closed. I know $k$ must to be infinite for sure since $\mathbb{A}^n$ is a finite set over a finite field $k$. Also, do we even consider $\mathbb{A}^n$ to be irreducible in that case i.e. $k$ is a finite field?
 A: The hypothesis $k$ algebraically closed is redundant. $k$ infinite is enough to get the result.
We can even prove the stronger result that an affine space $E$ over an infinite field $k$ can’t be the finite union of proper affine subspaces $E_1, \dots E_n$. Considering the underlying linear subspaces, it is enough to prove the similar result for linear subspaces.
For the proof we  denote $F_i= E_1 \cup \dots E_i$ and proceed by induction. The result is clear for $n=1$ as $E_1$ is supposed to be a proper linear subspace.
Suppose that $n \ge 2$ and assume by contradiction that $E=F_n$. According to induction hypothesis, $E\neq F_{n-1}$. So it exists $x \in E \setminus F_{n-1}$. Let also $y$ be in $E \setminus E_n$.
The map $u : k \to E$ defined by $u(\lambda)= \lambda x +y$ is taking values in $F_{n-1}$ as $\lambda x +y \in E_n$ implies $y \in E_n$. As $k$ is supposed to be infinite, it exists $\alpha \neq \beta$ and $j \le n-1$ such that $u(\alpha),u(\beta) \in E_j$. This leads to the contradiction
$$x =(\alpha-\beta)^{-1}(u(\alpha)-u(\beta)) \in E_j.$$
