Zariski closure in an irreducible component of a set

Question

Let $K$ be a field. We work with $K^n$ with Zariski topology. Let $A\subset K^n$ and let $V_1,\cdots,V_k$ be the irreducible components of the Zariski closure of $A$. Then $A\cap V_i$ is Zariski dense for all $i$. Why?

Answer 1

Recall that irreducible components of a closed set are closed, so we only need to prove $V_i\subseteq\overline{V_i\cap A}$. Since $\overline{X\cup Y}=\overline X\cup \overline Y$, we have that $V_i\subseteq \overline A=\overline {(V_1\cap A)}\cup\cdots\cup\overline{(V_k\cap A)}$. Since $V_i$ is an irreducible subspace and it is contained in a finite union of closed set, there must be some $j$ such that $V_i\subseteq\overline {V_j\cap A}$. Therefore, $V_i\subseteq V_j$. Since $V_1,\cdots, V_k$ is an irredundant family, $i=j$, which means that $V_i\subseteq\overline{V_i\cap A}$.