Closure of singleton point in Zariski topology

Question

Let $\{x\}$ denote the singleton corresponding to the prime ideal $p_x$ in the Zariski topology. I'm reading a proof that $cl(\{x\})=V(p_x)$, and it states "The closure $\{x\}$ is the intersection of all closed sets containing $x$. Furthermore, $x$ is in $V(E)$ if and only
if $E$ is a subset of $p_x$. Thus every closed set containing $x$ contains $V(p_x)$. ". I cannot understand how the first bold sentence implies the second bold one, can someone elaborate?

Answer 1

It is quite simple:

We have that $E\subset E'\implies V(E')\subset V(E)$.

So, if $x\in V(E)$, i.e. if $E\subset \mathfrak p_x$, we conclude instantly that $V(\mathfrak p_x)\subset V(E)$. Therefore, the intersection of all closed subsets containing $x$ is $V(\mathfrak p_x)$.