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.

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.

