Does there exist $m\ne n$ such that some open subset of $\mathbb A^m_k$ is homeomorphic with some open subset of $\mathbb A^n_k$?

Question

Let $k$ be an algebraically closed field. For every positive integer $n$ we can consider the affine $n$-space $\mathbb A^n_k$ with Zariski topology (the topology of zero sets of family of polynomials). As a topological space, $\mathbb A^n_k$ has krull dimension (maximal length of chain of closed irreducible subsets) $n$ , so for $m \ne n$ , $\mathbb A^m_k$ and $\mathbb A^n_k$ are not homeomorphic. My question is : Does there exist positive integers $m\ne n$ such that some open subset of $\mathbb A^m_k$ is homeomorphic with some open subset of $\mathbb A^n_k$ ?

Answer 1

No, any open subset of $\mathbb{A}^n_k$ is of dimension $n$. Use that the topology on $\mathbb{A}^n_k$ is generated by sets of the form $D(f)=\mathrm{Spec}(k[x_1,\ldots,x_n]_f)$. Now any maximal ideal in $k[x_1,\ldots,x_n]$ not containing $f$ gives rise to a chain of closed subsets of length $n$ in $D(f)$.