TITLE: Closure Operator and Set Operations
QUESTION [5 upvotes]: 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?

REPLY [2 votes]: 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.