$\mathbb A^n(k)$ and $\mathbb A^n(k)\setminus \{0\}$ are not homeomorphic

Let $k$ be an algebraic closed field. Why $\mathbb A^n(k)$ and $\mathbb A^n(k)\setminus\{0\}$ (for $n>1$) are not homeomorphic with respect to the Zariski topology?

Solution 1:

(This is not an answer, and everything that follows below was originally prompted by the fact that the O.P. did not exclude the case $n=1$ in his question when he first asked it, which he has now done at a later edit.)

Let $k$ be an algebraically closed field. Then the Zariski topology on $k$ is the finite complement topology, and similarly on $k- \{0\}$. Note that two spaces with finite complement topology are homeomorphic iff they have the space cardinality. In this case both $k$, $k - \{0\}$ have the same cardinality. So, they are in fact homeomorphic in the Zariski topology. In particular, I am not sure your statement is true.

What I wrote above may not illustrate just how weak the Zariski topology is. For a better example you can try to show that any two curves in $\mathbb{C}^2$, where by a curve I mean the zero set of an irreducible polynomial, are homeomorphic in the Zariski topology.

What is true though is that $k^n - \{0\}$ is not affine for $n \geq 2$. But, this question has been asked on MSE before. Just look for it. (Later addition: Note that as the user below pointed out, this does not imply that $k^n$ and $k^n - \{0\}$ are not homeomorphic. My comment about them not being affine was not meant as an argument for them not being homeomorphic.)

Here is a related mathoverflow link where Greg Kuperberg attempts to answer the question for $n = 2$ over $\mathbb{C}$ (thanks are completely due to another user for pointing this out to me who I cannot locate at the moment: https://mathoverflow.net/questions/78771/is-mathbbc2-homeomorphic-to-mathbbc2-0-0-with-the-zariski-topolog).

Finally, I asked some algebraic geometers at my institution, and they do not seem to be aware of this being a result.