Zariski sets dense in irreducible component of a Zariski closed 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 $A\setminus{\bigcup_{i\neq j} V_j}$ is Zariski dense in $V_i$ for all $i$. How can I show that?

Answer 1

$A\cap V_i$ is dense in $V_i$ and $U:=V_i\setminus\bigcup\limits_{i\ne j} V_j$ is a relatively open subset of $V_i$. Since $V_i$ is irreducible and $U\ne \emptyset$, $U$ is dense in $V_i$. Since $U$ is open, $A\setminus\bigcup\limits_{i\ne j} V_j=A\cap V_i\cap U$ is dense in $U$. Thus it's also dense in its closure in $V_i$, which is $V_i$ itself.