Prove that $\bigcup^{\infty}_{k=1}\overline{A_k} \subset \bar{B}$ if $B=\bigcup^{\infty}_{k=1}A_k $ Suppose $A_1,A_2,A_3,...$ are given subsets of $\mathbb{R} \space or \space \mathbb{R^2}$, and $B=\bigcup^{\infty}_{k=1}A_k. $
Prove that $\bigcup^{\infty}_{k=1}\overline{A_k} \subset \bar{B}$ , and give an example to show that the inclusion can be proper.
My attempt:
From what I understood from class, $\bar{A}$ is the closure of the set A,  $\bar{A} =A\cup A'$ where A' is the set of all limit points of A.
So I started with:
$\bar B=\overline{\bigcup^{\infty}_{k=1}A_k}=\bigcup^{\infty}_{k=1}\bar{A_k}$ since we proved in class that $\overline{A\cup B}=\bar{A}\cup \bar {B}$. I'm not sure we can generalize that to infinity though.
So clearly, $\bigcup^{\infty}_{k=1}\overline{A_k} \subset \bar{B}$ since they're equal.
But I'm pretty sure something is wrong. Any help/hint would really be appreciated.
thanks!!
 A: Hint: $\bigcup_{n=1}^\infty (0, 1-\frac1n) = (0,1)$, but $\bigcup_{n=1}^\infty \overline{(0, 1-\frac1n)} = [0,1) \ne [0,1] = \overline{(0,1)}$.
A: You certainly know that $A\subseteq B$ implies that $\overline A\subseteq\overline B$. Then since $A_j\subseteq\bigcup_{k=1}^\infty A_k$ for each $j$, you can conclude that $\overline {A_j}\subseteq\overline{\bigcup_{k=1}^\infty A_k}$, and therefore $\bigcup_{k=1}^\infty \overline {A_k}\subseteq\overline{\bigcup_{k=1}^\infty A_k}$. This works for any indexing set, not just the natural numbers.
But the equality in the case of two sets can be generalized to finitely many sets. We can even say more:
A family $(A_i)_{i\in I}$ is called locally finite if each point has a neighborhood intersecting only finitely many $A_i.$ In that case we can show that $(\overline{A_i})_I$ is also locally finite, and that $\overline{\bigcup_{i\in I}A_i}=\bigcup_{i\in I}\overline{A_i}$
In the example that @njguliyev gave, it does not hold since each neighborhood of the point $1$ intersects infinitely many sets $\left(0, 1-\frac1n\right).$
