# 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.

• Are you sure you copied the equation correctly? The first term on the right side of the $=$ isn't $\bigcup_{k=1}^\infty\overline{A_k}$? – bof Mar 7 '15 at 23:31
• Something definitely seems off; as it stands it's just saying the closure of an infinite union is the infinite union plus the closure of the "limiting set", which doesn't make sense necessarily. Isn't that long intersection of unions a "decreasing sequence of sets"? – jdods Mar 7 '15 at 23:37
• I have posted the problem correctly to my knowledge. And yes the long intersection of unions is a "decreasing sequence of sets" – Martin Mar 7 '15 at 23:45
• If you posted it correctly, then it was misstated (or is intended to be a ‘gotcha’), since the result is false, as noted in my answer. – Brian M. Scott Mar 7 '15 at 23:55
• @BrianM.Scott: I am still not able to close out this problem. I am trying to prove the corrected problem (as suggested by you). I have updated the problem statement with my attempt. I would appreciate if you can help me to close it out. – Martin Mar 9 '15 at 1:35

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 $\bigcup_{k\ge 1}A_k$.
• Thank You. This was helpful. – Martin Mar 8 '15 at 0:26
• @Martin: You're welcome. – Brian M. Scott Mar 8 '15 at 0:31
• Even after your explanation I am stuck with this problem. I have edited the problem to describe where I am stuck. I would appreciate your insight. – Martin Mar 8 '15 at 2:31
• thanks Brian. Can you help me with this question sir? math.stackexchange.com/questions/1179391/… – user203867 Mar 8 '15 at 2:46