(a) $\overline{B}$ is closed and contains $B$ , and, hence, $A$ , therefore, it contains the closure of $A$ .

(b) (c) Every closed set containing the union of $A_{\alpha}$ contains each set $A_{\alpha}$ , and, hence, its closure $\overline{A}_{\alpha}$ . Therefore, the closure of the union (the left hand side), which is the intersection of such closed sets, contains all $\overline{A}_{\alpha}$ and their union (the right hand side).

Now, the other direction. For any finite number of sets $\{A_{i}\}_{i=1}^{n}$ : $\overline{\cup_{i=1}^{n}A_{i}}=$ $\cap_{E\supset(\cup_{i=1}^{n}A_{i}),E\mbox{ is closed}}E=$ $\cap_{\{C_{i}\supset A_{i},C_{i}\mbox{ is closed}\}_{i=1}^{n}}(\cup_{i=1}^{n}C_{i})=$ $\cup_{i}\overline{A}_{i}$ (in fact, this is a proof for both directions in the case of a finite number of sets). For infinite number of sets, some collections of closed sets $C_{i}$ have non-closed unions so that the intersection over all closed sets $E$ is the intersection over a subcollection of sets $\{C_{i}\}_{i=1}^{n}$ (which makes the right hand side smaller, as we intersect over a larger collection of sets). This suggests that the right hand side is a proper subset of the left hand side if there is a collection of closed sets $C_{\alpha}$ containing $A_{\alpha}$ , respectively, such that their union is not closed, and there is no closed set contained in the union that would contain all $A_{\alpha}$ . For example, if we take $A_{n}=\{\frac{1}{n}\}$ and $A=\cup_{n\in\mathbb{Z}_{+}}A_{n}$ then $\overline{A}=A\cup\{0\}\supsetneq\cup_{n}\overline{A}_{n}=A$ (here, for example, $C_{n}=[\tfrac{1}{2n},2]$ is closed and contains $A_{n}$ , but the union $\cup_{n\in\mathbb{Z}_{+}}C_{n}=(0,2]$ is not closed, and there is no closed $E$ such that $A\subset E\subset(0,2]$ , so the right hand side has an “extra” intersection with $(0,2]$ which the left hand side does not have).