Prove that $\overline {B^*} = B.$ 

Let $R$ be a commutative ring with identity. Let $\operatorname {Max} (R)$ denote the set of all maximal ideals of $R$ and let $\operatorname {J} (R)$ denote the set of all prime ideals of $R$ which can be written as the intersection of maximal ideals of $R.$ Let $B \subseteq \operatorname {J} (R)$ be a closed subset of $\operatorname {J} (R).$  Let $B^*= B \cap \operatorname {Max} (R).$ Let $\overline {B^*}$ denote the closure of $B^*$ in $\operatorname {J} (R).$ Then show that $\overline {B^*} = B.$


My attempt $:$
Let $\mathcal {I} (B)$ denote the vanishing ideal of $B.$ Then $$\mathcal {I} (B) = \bigcap\limits_{\substack {p \in B \\ \text {p prime}}} p.$$
Now what is $\mathcal {I} (B^*)$? $$\mathcal {I} (B^*) = \bigcap\limits_{m \in B^*} m.$$
Clearly $\mathcal {I} (B) \subseteq \mathcal {I} (B^*).$ If we can prove that the other part is also equally valid we are through. Because let $V(I)$ denote the zero set of the ideal $I$ of $R.$ Then we know that for any $A \subseteq \operatorname {J} (R)$ $$V(\mathcal {I} (A)) = \overline {A}.$$
In this case if we can prove that $\mathcal {I} (B) = \mathcal {I} (B^*)$ then we have $$\begin{align*} V(\mathcal {I} (B)) & = V(\mathcal {I} (B^*)) \implies \overline {B^*} = \overline {B}. \end{align*}$$
Now since $B$ is closed in $\operatorname {J} (R)$ we have $\overline {B} = B.$ This shows that $$\overline {B^*} = B$$ as required. 
But how do I prove that $\mathcal {I} (B^*) \subseteq \mathcal {I} (B)$? Please help me in proving this. Any help will be highly appreciated.
Thank you very much for your valuable time.
 A: $$\mathcal I(B)= \bigcap_{p\in B}p$$
$$\mathcal I(B^*)= \bigcap_{m\in B\cap\mathrm{Max}(R)} m$$
$B$ is closed in $J(R)$, so if $p\in B$ and $p\subset m \in\mathrm{Max}(R)$ we have $m\in B^*$.
In other words $$B\cap\mathrm{Max}(R)\supset \{ m\in\mathrm{Max}(R)\mid \exists p\in B\text{ s.t. } p\subset m \} $$ and the other inclusion is trivial.
Thus we have $$\mathcal I(B)=\bigcap_{p\in B} \bigcap_{p\subset m \in \mathrm{Max}(R)} m= \bigcap _{m \in \mathrm{Max}(R)\cap B} m= \mathcal I(B^*).$$
A: I have found an answer with much simpler argument.
Since $B$ is closed in $\operatorname {J} (R)$ so $\exists$ an ideal $I$ of $R$ such that $$B = V(I) = \{p \in \operatorname {J} (R) : p \supseteq I \}.$$
Let $q \in  B.$ Then since $B \subseteq \operatorname {J} (R)$ so $\exists$ an indexing set $\Lambda$ such that $$q = \bigcap\limits_{\lambda \in \Lambda} m_{\lambda}$$ where each $m_{\lambda}$ is maximal. So clearly $m_{\lambda} \in \operatorname {J} (R)$ containing $I,$ for all $\lambda \in \Lambda.$ So $m_{\lambda} \in B,$ for all $\lambda \in \Lambda.$ Also since $m_{\lambda} \in \operatorname {Max} (R)$ for each $\lambda \in \Lambda$ we have $m_{\lambda} \in B^*,$ for each $\lambda \in \Lambda.$ This shows that $$\{m_{\lambda} : \lambda \in \Lambda \} \subseteq B^*.$$ Therefore we have $$\bigcap\limits_{m \in B^*} m \subseteq \bigcap\limits_{\lambda \in \Lambda} m_{\lambda} = q.$$ Since $q \in B$ was arbitrarily taken so we have $$\bigcap\limits_{m \in B^*} m \subseteq \bigcap\limits_{p \in B} p.$$ This shows that $$\mathcal {I} (B^*) \subseteq \mathcal {I} (B)$$ as required.
