Prove that the closure of a set equals the intersection of all closed sets containing it. I do think that the answer is easy but I need help in writing a proper proof.
Let $(E,d)$ be a metric space, and $A,B \subset E$ then:
Prove $\bar A$ coincides with the intersection of the set of all subsets $F \subset E$ such that $F$ contains $A$ and $F$ is closed set in $E.$
 A: Let $B$ be the intersection of all closed sets in $E$ that contain the set $A.$ Then, as $\bar A$ is closed and contains $A,$ it follows that $B \subset \bar A.$ For the reverse, if $x$ belong to the closure of $A$ in $E$ and $F$ is a closed set in $E$ that contains $A,$ then for every $r > 0,$ the ball $B(x, r)$ intersects $A$ and therefore, $F$ too, hence $x \in B.$ QED
A: Let $A \subset E$ and let $F_\alpha$ be closed in E for each $\alpha$ in a non-empty index set I.
Let $A \subset F_\alpha \,\, \forall \alpha \in I$
Let $\bar{A}$ be the closure of A in E. Since $\bar{A} \subset F_\alpha \,\, \forall \alpha \in I$ then $\bar{A} \cap F_\alpha = \bar{A} \,\, \forall \alpha \in I$. This implies $(\bar{A} \cap F_{\alpha_1}) \cap F_{\alpha_2} = \bar{A} \cap F_{\alpha_2} = \bar{A}$. Continuing inductivily we get $\bar{A} \cap (\cap_{\alpha \in I} F_\alpha) = \bar{A}$. Since for any set $D\subset E \,\, D \cap D = D$ then we have $\bar{A} = \cap_{\alpha \in I} F_\alpha$. Q.E.D
Note: That $\bar{A} \subset F_\alpha \,\, \forall \alpha \in I$ with $A \subset F_\alpha \,\, \forall \alpha \in I$ is proved in Dangello&Seyfried, ${Introductory Real Analysis}$ on p. 248, Thm 11.3.
