Geometric Interpretation of Jacobson rings -- Every locally closed subsets of $\operatorname{Spec} A$ consists of a single point is closed I'm currently working on Atiyah&MacDonald's book on commutative algebra. I'm trying on Exercise 5.26, and remaining to show that

Question: $A$ is a Jacobson ring if and only if
$$
\text{every locally closed subsets of } \operatorname{Spec} A \text{ consists of a single point is closed} \quad (\star)
$$
Here by saying $A$ is Jacobson, by Exercise 5.23, we mean every non-maximal prime ideal in $A$ is equal to the intersection of the prime ideals which contain it strictly.

My attempts: Let $\{ \mathfrak{p}\}$ be a singleton in $\operatorname{Spec} A$. Then it is locally closed if and only if $\{ \mathfrak{p}\}$ is open in its closure $V(\mathfrak{p})$. Hence the $(\star)$ is equivalent to say that for any prime ideal of $\mathfrak{p}$ of $A$, if $\{ \mathfrak{p}\}$ is open in $V(\mathfrak{p})$, then $\mathfrak{p}$ is a maximal ideal of $A$. Notice that conversely for maximal ideals $\mathfrak{m}$, $\mathfrak{m}$ is indeed open in $V(\mathfrak{m})$. So actually, $(\star)$ is equivalent to that
$$
\forall \mathfrak{p} \in \operatorname{Spec} A, ( \{\mathfrak{p}\} \text{ is 
 NOT open in } V(\mathfrak{p}) \Leftrightarrow \mathfrak{p} \text{ is not a maximal ideal of } A ).
$$
So recall the definition of Jacobson rings quoted above, we need to show that for all $\mathfrak{p} \in \mathrm{Spec} A$
$$
\{\mathfrak{p}\} \text{ is NOT open in } V(\mathfrak{p}) \Leftrightarrow \mathfrak{p} = \bigcap_{\mathfrak{p}^{\prime} \in V(\mathfrak{p}) - \{\mathfrak{p}\}} \mathfrak{p}^{\prime}. \quad\quad (\star\star).
$$
So my question actually is: how to show the above $(\star\star)$. I got stuck here and actually having a hard time describing open sets of $\operatorname{Spec} A$, after trying on principal opens and failed.
Thank you all for commenting and answering! :)
 A: I don't follow the last two equivalences in your question. You're correct that $(\star)$ is equivalent to: the prime ideal $\mathfrak{p}$ is maximal if and only if $\{\mathfrak{p}\}$ is open in $V(\mathfrak{p})$. But then you say that you want to show that $\mathfrak{p}$ is not maximal if and only if $\{\mathfrak{p}\}$ is open in $\mathrm{Spec}(A)$. This is strange, since the negation of " $\{\mathfrak{p}\}$ is open in $V(\mathfrak{p})$" is not "$\{\mathfrak{p}\}$ is open in $\mathrm{Spec}(A)$"!
Anyway, here is a solution to the exercise:
Suppose $A$ is a Jacobson ring. Let $\{\mathfrak{p}\}$ be a singleton which is locally closed in $\mathrm{Spec}(A)$. Then $\{\mathfrak{p}\}$ is open in $V(\mathfrak{p})$, so there is a basic open set $D(f)$ such that $D(f)\cap V(\mathfrak{p}) = \{\mathfrak{p}\}$. That is, $f\in A$, $f\notin \mathfrak{p}$, and for all prime ideals $\mathfrak{q}$ with $\mathfrak{p}\subsetneq \mathfrak{q}\subseteq A$, $f\in \mathfrak{q}$. It follows that $$\mathfrak{p}\neq \bigcap_{\mathfrak{p}\subsetneq \mathfrak{q}}\mathfrak{q},$$ since $f$ is in the right-hand-side but not the left-hand-side. By your characterization of Jacobson rings, $\mathfrak{p}$ is maximal. So $\{\mathfrak{p}\}$ is closed.
Conversely, suppose every singleton which is locally closed in $\mathrm{Spec}(A)$ is closed. Let $\mathfrak{p}$ be a non-maximal prime ideal in $A$. Then $\{\mathfrak{p}\}$ is not closed, so it is not open in its closure $V(\mathfrak{p})$. We have $$\mathfrak{p}\subseteq \bigcap_{\mathfrak{p}\subsetneq \mathfrak{q}}\mathfrak{q}.$$ Suppose for contradiction that the reverse inclusion is not true. Then there is some $f\in \bigcap_{\mathfrak{p}\subsetneq \mathfrak{q}}\mathfrak{q}$ such that $f\notin \mathfrak{p}$. Now $D(f)\cap V(\mathfrak{p}) = \{p\}$, so $\mathfrak{p}$ is open in $V(\mathfrak{p})$, contradiction. Thus, by your characterization of Jacobson rings, $A$ is Jacobson.
