Let $X\subsetneq \mathbb A^n$ be Zariski closed and countable. Is $X$ then finite?
If $n=1$, then its clear, because the Zariski closed sets in $\mathbb A^1$ are just the zero set of some polynomial, so every Zariski closed set is finite.
But what about higher dimensions?
Definition of $\mathbb A^n$:
We define affine $n$-space over $k$,
denoted $\mathbb A^n$, to be the set of all $n$-tuples of elements of $k$.
Edit: Parthiv Basu already gave a valid answer to this question.
However, I forgot to mentaion that I want to work over an algebraically closed field $k$. Actually, I had $k = \mathbb C$ in mind. I posted this as a new question
If you are considering the affine $n$-space over some field $\mathbb{k}$ then it is certainly true that the closed subsets in $\mathbb{A}^1(\mathbb{k})$ are just $\emptyset$, finite subsets of $\mathbb{A}^1(\mathbb{k})$ and $\mathbb{A}^1(\mathbb{k})$ itself. But for $\mathbb{A}^2(\mathbb{Q})$, the unit circle $V(x^2 + y^2-1)$ is a closed countable set but not finite.