Unions of Sets of Limit Points So I am trying to answer the question #2.7 from Rudin's Principles of Math Analysis. The problem is as follows: 
Some notation: $\bar{E}$ is the closure of $E$ and $E'$ is the set of limit point of $E$. (I don't know if this is standard notation)
Let $A_1,A_2, ..., A_n$ ,for $n \in \mathbb{N}$, be subsets of a metric space. 
If $B_n = \bigcup_{i=1}^{n} A_i$ prove that $\bigcup_{i=1}^{n} \bar{A}_i = \bar{B}$. 
What I have so far: Suppose that $A_1,A_2, ..., A_n$ for $n \in \mathbb{N}$ be subsets of a metric space and $B_n = \bigcup_{i=1}^{n} A_i$. Consider $\bigcup_{i=1}^{n} \bar{A}_i$, by definition of the closure, $\bigcup_{i=1}^{n} \bar{A}_i = \bigcup_{i=1}^{n} (A_i \cup A^{'}_i)$ which we can write as $\bigcup_{i=1}^{n} A_i \cup \bigcup_{i=1}^{n} A_i^{'}$ then by our assumption, $\bigcup_{i=1}^{n} A_i \cup \bigcup_{i=1}^{n} A_i^{'} = B_n \cup \bigcup_{i=1}^{n} A_i^{'}$
Assuming that I am correct so far, then where I am trying to go is to say $\bigcup_{i=1}^{n} A_i^{'} = (\bigcup_{i=1}^{n} A_i)^{'}$ then make the same substitution I did previously. So my question is that is this true: $\bigcup_{i=1}^{n} A_i^{'} = (\bigcup_{i=1}^{n} A_i)^{'}$ intuitively it feels true, but I wouldn't be quite sure how to prove this or maybe this is just an obvious statement? Any guidance or tips would be highly appreciated!
 A: The idea is to prove that $(\bigcup_{k=1}^n A_i)' \subseteq \bigcup_{k=1}^n A_i'$. To this end, suppose it weren't; then there would be $x$ a limit point of $\bigcup_{k=1}^n A_i$, but not any of the $A_i$. But in this case there would for each $k$ be an $\epsilon_k$ such that $|x - \alpha| > \epsilon_k$ for every $\alpha \in A_k$. Then taking the minimum of these $\epsilon_k$ produces an $\epsilon$ such that $|x - \alpha| > \epsilon$ for every $\alpha \in \bigcup_{k=1}^n A_i$, contradicting $x$ being a limit point.
The other direction, $(\bigcup_{k=1}^n A_i)' \supseteq \bigcup_{k=1}^n A_i'$, should be a fairly straightforward exercise, which I will leave to you. Together these prove the claim!
A: It is true.
If $x$ is a limit point of some $A_k$, then it is obviously a limit point of the larger set $\cup_i A_i$. Hence, $\bigcup_{i=1}^{n} A_i^{'} \subseteq (\bigcup_{i=1}^{n} A_i)^{'}$ is proved.
If $x$ is a limit point of $\cup_i A_i$, then there is a sequence of points $x_j\to x$ where each $x_j$ lies in $A_k$ for some $k$. Since there are only finitely many $A_i$'s, some $k$ must appear infinitely many times. For this $A_k$, $x\in A_k'$. This proves $\bigcup_{i=1}^{n} A_i^{'} \supseteq (\bigcup_{i=1}^{n} A_i)^{'}$.
