Closure Operator and Set Operations In Engelking, General Topology stand the following exercise:
Show that for any sequence $A_1, A_2, \ldots$ of subsets of a topological space we have
$$
 \overline{\bigcup_{i=1}^{\infty} A_i} = \bigcup_{i=1}^{\infty} \overline{A_i} \cup \bigcap_{i=1}^{\infty} \overline{\bigcup_{j=0}^{\infty} A_{i+j}}.
$$
I have no idea how to show this, I can't even show that the right side yields a closed set, because when I build the complement I got infinite intersections, which aren't necessarily open. Any hints how to solve this exersice?
 A: $\newcommand{\cl}{\operatorname{cl}}$For each $i\in\Bbb Z^+$
$$\cl\bigcup_{j\ge 0}A_{i+j}\subseteq\cl\bigcup_{j\ge 1}A_j\;,$$
so 
$$\bigcap_{i\ge 1}\cl\bigcup_{j\ge 0}A_{i+j}\subseteq\cl\bigcup_{i\ge 1}A_i\;,$$
and therefore
$$\cl\bigcup_{i\ge 1}A_i\supseteq\bigcup_{i\ge 1}\cl A_i\cup\bigcap_{j\ge 1}\cl\bigcup_{i\ge 0}A_{i+j}\;.$$
Suppose that 
$$x\in\left(\cl\bigcup_{i\ge 1}A_i\right)\setminus\bigcup_{i\ge 1}\cl A_i\;.$$
For any $n\in\Bbb Z^+$ we know that
$$\cl\bigcup_{i=1}^nA_k=\bigcup_{i=1}^n\cl A_i\;,$$
so for each $n\in\Bbb Z^+$ we have
$$x\in\left(\cl\bigcup_{i\ge 1}A_i\right)\setminus\cl\bigcup_{i=1}^nA_i=\left(\cl\bigcup_{i\ge n+1}A_i\cup\cl\bigcup_{i=1}^nA_i\right)\setminus\cl\bigcup_{i=1}^nA_i\subseteq\cl\bigcup_{i\ge n+1}A_i\;;$$
do you see that this implies that 
$$x\in\bigcap_{j\ge 1}\cl\bigcup_{i\ge 0}A_{i+j}$$
and hence that 
$$\cl\bigcup_{i\ge 1}A_i\subseteq\bigcup_{i\ge 1}\cl A_i\cup\bigcap_{j\ge 1}\cl\bigcup_{i\ge 0}A_{i+j}\;?$$
A: The inclusion (RHS) $\subseteq$ (LHS) is trivial, so I will give you a hint for the reverse inclusion:
Assume that $x$ is not in the right hand side. Then there is a $j\ge1$ such that $x\notin\overline{\bigcup_{i=j}^\infty A_j}$, i.e. there is a neighborhood $V$ disjoint from this union starting at $j$. But for each $i\ge0$ there is also a neighborhoods $U_i$ disjoint from $A_i$. Now intersect $W:=V\cap U_1\cap...\cap U_j$. Does this $W$ intersect the union of all $A_i$?
A: I am still looking through your solutions, but meanwhile I came up with another for the direction (LHS) $\subseteq$ (RHS). Can you please look if its valid?
Let $x \in \overline{\bigcup_{i=1}^{\infty} A_i}$, suppose $x \notin \bigcup_{i=1}^{\infty}\overline{A}_i$ (otherwise we are finished), then we have to show for all $i$ that $x \in \overline{\bigcup_{j\ge i} A_j}$. Suppose there is some $i$ for which $x \notin \overline{\bigcup_{j\ge i} A_j}$, then as $x \in \overline{\bigcup_{i=1}^{\infty} A_i} = \overline{\bigcup_{j=1}^{i-1} A_i} \cup \overline{\bigcup_{j\ge i} A_i}$. It must be that $x \in \overline{\bigcup_{j=1}^{i-1} A_i}$, but because $x \notin \bigcup_{i=1}^{\infty}\overline{A}_i$ and $\overline{\bigcup_{j=1}^{i-1} A_i} = \bigcup_{j=1}^{i-1} \overline{A_i} \subseteq \bigcup_{i=1}^{\infty} \overline{A}_i$ we have a contradiction.
