Identity of closure of sets in a topological space 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.

Let $\displaystyle x\in\overline{\bigcup_{i\in\mathbb{N}^*}A_i}$. We assume that $\displaystyle  x\notin\bigcup_{i\in\mathbb N^*}\overline{A_i}$ (if it's not the case we are done).We have to show that for all $i\geq 1$ we have $\displaystyle x\in\overline{\bigcup_{j\geq i}A_j}$. Let $V$ a neighborhood of $x$. Since $\displaystyle  x\notin\bigcup_{i\in\mathbb N^*}\overline{A_i}$ we can find for all $i\geq 1$ an open set $U_i$ which contains $x$ and such that $U_i\cap A_i=\emptyset$. For $i\geq 1$, let $\displaystyle V_i=\bigcap_{k=1}^iU_k$: this is an open set which contains $x$. Since $V\cap V_i$ is still a neighborhood of $x$ and $\displaystyle x\in\overline{\bigcup_{i\in\mathbb{N}^*}A_i}$ we have $\displaystyle V\cap V_i\cap\bigcup_{k\in\mathbb{N}}A_k \neq \emptyset$. Now we can conclude since $\displaystyle V_i\cap\bigcup_{k\in\mathbb{N}}A_k =V_i\cap \bigcup_{k\geq i+1}A_k$.
