If $A_1,A_2,...$ is a sequence of subsets of a topological space. Prove following If $A_1,A_2,...$ is a sequence of subsets of a topological space. Prove
$\overline{\bigcup_{k=1}^{\infty}A_k} = \bigcup_{k=1}^{\infty}A_k \cup \bigcap_{k=1}^{\infty}\bigg(\overline{\bigcup_{l=0}^{\infty}A_{k+l}}\bigg)$
I am first trying to decipher the right hand side of the equation
Let $x\in$ RHS $\Rightarrow $ $x\in\bigcap_{k=1}^{\infty}A_k$ and $x\in\bigcap_{k=1}^{\infty}\bigg(\overline{\bigcup_{l=0}^{\infty}A_{k+l}}\bigg)$
Now $x\in\bigcap_{k=1}^{\infty}\bigg(\overline{\bigcup_{l=0}^{\infty}A_{k+l}}\bigg)$ $\Rightarrow $ $x\in \bigcap_{k=1}^{\infty}\overline{ \left \{ A_{k+0} \cup A_{k+1} \cup A_{k+2} \cup A_{k+3} ... \right \}} $
$\Rightarrow$ $x\in \overline{(A_1 \cup A_2 \cup A_3...)} \cap\overline{(A_2 \cup A_3 \cup A_4...)} \cap \overline{(A_3 \cup A_4 \cup A_5...)}\cap.... $
But this is not taking me anywhere. I would appreciate if someone can point me in right direction.
@Brian: Here is how I am trying to attempt the corrected problem.. As per your comment the correct problem should be 
$\overline{\bigcup_{k=1}^{\infty}A_k} = \bigcup_{k=1}^{\infty}\overline{A_k} \cup \bigcap_{k=1}^{\infty}\bigg(\overline{\bigcup_{l=0}^{\infty}A_{k+l}}\bigg)$
Let $x\in \overline{\bigcup_{k=1}^{\infty}A_k}$ this implies that every nbhd $U$ of $x$ will intersect with some $A_k$ where $k\geq 1$. 
To prove in forward direction $x\in \bigcup_{k=1}^{\infty}\overline{A_k}$ implies that $x$ belongs to the closure of at least one of the $A_k$ where $k\geq 1$.Thus every neighborhood of $x$ intersects with at least one perticular $A_k$ where $k\geq 1$.
I believe that this would be good enough to imply forward inclusion that is LHS $\subset$ RHS. Please let me know if it does not.
Conversely Let $x\in \bigcup_{k=1}^{\infty}\overline{A_k} \cup \bigcap_{k=1}^{\infty}\bigg(\overline{\bigcup_{l=0}^{\infty}A_{k+l}}\bigg)$ 
As you explained $x\in \bigcap_{k=1}^{\infty}\bigg(\overline{\bigcup_{l=0}^{\infty}A_{k+l}}\bigg)$ implies that  for each $k\geq1$ and each nbhd $U$ of $x$ there exist an $l\geq k$ such that $(U\cap A_l)\ne\varnothing $. Thus for each $k\geq1$ every nbhd of $x$ intersects with some $A_l$ (where $l\geq k)$. Also as mentioned before $x\in \bigcup_{k=1}^{\infty}\overline{A_k}$ implies that $x$ belongs to the closure of at least one of the $A_k$ where $k\geq 1$. I am having hard time in using these two deductions to imply that $x\in$ LHS. I would appreciate if you can help me.
 A: The right-hand side is
$$\bigcup_{k\ge 1}A_k\cup\bigcap_{k\ge 1}\left(\operatorname{cl}\bigcup_{\ell\ge k}A_\ell\right)\;;\tag{1}$$
$x$ is in this set if and only if $x\in\bigcup_{k\ge 1}A_k$ or $x\in\bigcap_{k\ge 1}\left(\operatorname{cl}\bigcup_{\ell\ge k}A_\ell\right)$. The first alternative is clear enough, but the second takes some thought to untangle.
First note that as $k$ gets bigger, $\bigcup_{\le\ge k}A_\ell$ gets smaller: sets $A_n$ with $n$ smaller than $k$ are dropping out of the union. Thus, $\operatorname{cl}\bigcup_{\ell\ge k}A_\ell$ is getting smaller as well. And in order for $x$ to be in $\bigcap_{k\ge 1}\left(\operatorname{cl}\bigcup_{\ell\ge k}A_\ell\right)$, $x$ must be in each of the sets $\operatorname{cl}\bigcup_{\ell\ge k}A_\ell$ with $k\ge 1$. 
Suppose that $x\in\operatorname{cl}\bigcup_{\ell\ge k}A_\ell$; then every open nbhd of $x$ intersects $\bigcup_{\ell\ge k}A_\ell$. That is, if $U$ is an open nbhd of $x$, then there is an $\ell\ge k$ such that $U\cap A_\ell\ne\varnothing$. Thus,
$$x\in\bigcap_{k\ge 1}\left(\operatorname{cl}\bigcup_{\ell\ge k}A_\ell\right)$$
if and only if for each $k\ge 1$ and each open nbhd $U$ of $x$ there is an $\ell\ge k$ such that $U\cap A_\ell\ne\varnothing$. Another, perhaps more intuitive way to say this is that each open nbhd of $x$ intersects infinitely many of the sets $A_n$. Informally, then, you’re asked to prove that $x$ is in the closure of $\bigcup_{k\ge 1}A_k$ if and only if either $x$ is in the union itself, or every open nbhd of $x$ intersects infinitely many of the sets $A_k$.
This is false. For $k\in\Bbb Z^+$ let $A_k=\left(\frac1{k+1},\frac1k\right)$; then $\bigcup_{k\ge 1}A_k=(0,1)\setminus\left\{\frac1k:k\ge 2\right\}$, and its closure is $[0,1]$. In particular, $\frac12$ is in the closure. However, 
$$\frac12\notin\left[0,\frac13\right]=\operatorname{cl}\bigcup_{\ell\ge 3}A_\ell\;,$$
so 
$$\frac12\notin\bigcup_{k\ge 1}A_k\cup\bigcap_{k\ge 1}\left(\operatorname{cl}\bigcup_{\ell\ge k}A_\ell\right)\;.$$
The statement becomes true if you replace $\bigcup_{k\ge 1}A_k$ in $(1)$ with $\bigcup_{k\ge 1}\operatorname{cl}A_k$, and the proof isn’t too hard if you think about $\bigcap_{k\ge 1}\left(\operatorname{cl}\bigcup_{\ell\ge k}A_\ell\right)$ the way that I suggested above.
Added in response to edited question: Oddly enough, it’s showing that the right-hand side is contained in the left-hand side that’s more straightforward, at least if approached properly. If 
$$x\in\bigcup_{k\ge 1}\operatorname{cl}A_k\;,$$
then $x\in\operatorname{cl}A_\ell$ for some $\ell\in\Bbb Z^+$, and $A_\ell\subseteq\bigcup_{k\ge 1}A_k$, so $x\in\operatorname{cl}A_\ell\subseteq\operatorname{cl}\bigcup_{k\ge 1}A_k$. And if
$$x\in\bigcap_{k\ge 1}\left(\operatorname{cl}\bigcup_{\ell\ge k}A_\ell\right)\;,$$
then in particular $x\in\operatorname{cl}\bigcup_{\ell\ge 1}A_\ell$. This shows that the right-hand side is a subset of the left-hand side.
To show the other inclusion, suppose that $x\in\operatorname{cl}\bigcup_{k\ge 1}A_k$. If $x\in\bigcup_{k\ge 1}\operatorname{cl}A_k$, we’re done, so suppose not; we want to show that $x\in\bigcap_{k\ge 1}\left(\operatorname{cl}\bigcup_{\ell\ge k}A_\ell\right)$, i.e., that every open nbhd of $x$ intersects infinitely many of the sets $A_k$. 
Let $U$ be an open nbhd of $x$. Suppose that $\{k\in\Bbb Z^+:U\cap A_k\ne\varnothing\}$ is finite; then it has a maximum element $m$.


*

*Derive a contradiction by showing that $U\setminus\bigcup_{k=1}^m\operatorname{cl}A_k$ is an open nbhd of $x$ disjoint from $\bigcup_{k\ge 1}A_k$.

