Proving that the union of the limit points of sets is equal to the limit points of the union of the sets My instructor gave us an exercise to do, to show that this equality holds:
$$\bigcup_{k=1}^m A_k'=\left(\bigcup_{k=1}^m A_k\right)^{\!\prime}.$$
My thoughts on approaching this question is by contradiction, suppose that there is an element that satisfies the left side of the equality and not the right side. Then there is a deleted ball with radius $r$ such that $B(x,r)$ intersect the left side of the equality is non empty while there is a deleted ball with radius $r$ such that $B(x,r)$ intersect the right side of the equality is empty. Is this the right approach in proving this? This is as far as I can go.
Thanks for the help in advance.
 A: Hint: Try doing it with two sets and then proceed by induction.
Detail on reverse inclusion: Suppose $x\notin A'\cup B'$. Then there is an open set $U$ about $x$ such that $A\cap U\subseteq\{x\}$ and there is an open set $V$ about $x$ such that $B\cap V\subseteq\{x\}$. Then $U\cap V$ is an open set about $x$ and $(U\cap V)\cap (A\cup B)\subseteq \{x\}$, so $x\notin (A\cup B)'$.
A: It’s very easy to show that $A_\ell'\subseteq\left(\bigcup_{k=1}^mA_k\right)'$ for each $\ell\in\{1,\ldots,m\}$ and hence that
$$\bigcup_{k=1}^mA_k'\subseteq\left(\bigcup_{k=1}^mA_k\right)'\;;$$
the harder part is showing that
$$\left(\bigcup_{k=1}^mA_k\right)'\subseteq\bigcup_{k=1}^mA_k'\;.\tag{1}$$
Proof by contradiction is a reasonable guess, but it’s not needed. 
Suppose that $x\in\left(\bigcup_{k=1}^mA_k\right)'$; then for each $n\in\Bbb Z^+$ there is an $$x_n\in B\left(x,\frac1n\right)\cap\bigcup_{k=1}^mA_k\;,$$ and it’s clear that the sequence $\langle x_n:n\in\Bbb Z^+\rangle$ converges to $x$. For each $n\in\Bbb Z^+$ there is a $k_n\in\{1,\ldots,m\}$ such that $x_n\in A_{k_n}$.


*

*Show that there is an $\ell\in\{1,\ldots,m\}$ such that $\{n\in\Bbb Z^+:k_n=\ell\}$ is infinite. Let $L=\{n\in\Bbb Z^+:k_n=\ell\}$.  

*Show that $\langle n:n\in L\rangle$ is a sequence in $A_\ell$ converging to $x$.  

*Conclude that $(1)$ must hold.

A: A slightly different way to prove
$\left(\bigcup_k A_k\right)'\subseteq \bigcup_k A_k'$
is as followed.
Say $p\in\left(\bigcup_k A_k'\right)^c$, i.e. $p$ is not a limit point of any of $A_k$, there exists a neighborhood $N(p)$ (a ball of radius $r$ centered at $p$, if you will) that does not intersect any of $A_k$.
This means $N(p)$ does not intersect $\bigcup_k A_k$ either, so therefore, $p$ cannot be a limit point of $\bigcup_k A_k$, i.e. $p\in\left(\left(\bigcup_k A_k\right)'\right)^c$.
As a result,
$\left(\bigcup_k A_k'\right)^c \subseteq \left(\left(\bigcup_k A_k\right)'\right)^c,$
so
$\left(\bigcup_k A_k\right)' \subseteq \bigcup_k A_k'$
