A fun problem: show that for any sequence $A_{1},A_{2},...$ of subsets of a topological space we have:
$\overline{\bigcup_{i=1}^{\infty} A_{i}} = \bigcup_{i=1}^{\infty} \overline{A_{i}} \cup \bigcap_{i=1}^{\infty} \overline{\bigcup_{j=0}^{\infty}A_{i+j}}.$
So I think I have the inclusion $\supseteq$. Let's see:
First note that for each $i$ we have $\displaystyle A_{i} \subseteq \bigcup_{i=1}^{\infty} A_{i}$. Taking closure on both sides gives:
$\overline{A_{i}} \subseteq \overline{\bigcup_{i=1}^{\infty} A_{i}}.$
Taking the union on both sides from $i=1$ to $\infty$ yields:
$\bigcup_{i=1}^{\infty} \overline{A_{i}} \subseteq \overline{\bigcup_{i=1}^{\infty} A_{i}}.$ Call this (*). Now note that (taking $i=1$)
$\bigcap_{i=1}^{\infty} \overline{\bigcup_{j=0}^{\infty}A_{i+j}} \subseteq \overline{\bigcup_{j=0}^{\infty} A_{j+1}}.$
Thus:
$\bigcup_{i=1}^{\infty} \overline{A_{i}} \cup \bigcap_{i=1}^{\infty} \overline{\bigcup_{j=0}^{\infty}A_{i+j}} \subseteq \bigcup_{i=1}^{\infty} \overline{A_{i}} \cup \overline{\bigcup_{j=0}^{\infty} A_{j+1}}.$
Now by (*) it follows that the above set is a subset of:
$\overline{\bigcup_{i=1}^{\infty} A_{i}} \cup \overline{\bigcup_{j=0}^{\infty} A_{j+1}}.$
But the latter set is equal to $\displaystyle \overline{\bigcup_{i=1}^{\infty} A_{i}}$, so we have the inclusion $\supseteq$.
Is this OK? Now, how to prove the other inclusion? I tried contradiction but gets messy.