intersection of all neighborhoods of a point in zariski topology.

Question

Consider the affine space $A^n$ with the Zariski topology, V a variety (with the induced topology) and let $P\in V$ a point. Let B be the set of all neighborhoods of the point P in V. Is it true that $P=\bigcap\limits_{U_i\in B} U_i$?

Obviously this is not true on a general topological space (e.g consider the trivial topology where the only open sets are $\emptyset$ and the whole space), so the fact that we have Zariski topology is important.

Apart from that, I' m not sure on how to proceed. Intuition tells me that this is indeed true, but given the fact that open sets here are very big in size (they 're dense, so no two open sets can have an empty intersection), I 'm not confident of the result. Any hint would be welcome.

Answer 1

$(a_1,...,a_n)=V(X-a_1,...,X-a_n)$, this implies that $U(a_1,...,a_n)=A^n-(a_1,..,a_n)$ is open.

$(a_1,..,a_n)=\cap U(b_1,...,b_n), (b_1,..,b_n)\neq (a_1,..,a_n)$.

Answer 2

Follows from the fact that $\Bbb A^n$ is $T_1$ and that subspace of $T_1$ space is $T_1$.