For each $\gamma$, the derived set of $A_\gamma$ is closed. Show that the derived set of $\cup A_\gamma$ is closed. 
Let $(X,\tau)$ be a topological space, and $\{A_\gamma\}_{\gamma \in \Gamma}$ are subsets of $X$. If for each $\gamma \in \Gamma$, the derived set of $A_\gamma$ is closed, show that the derived set of $\cup_{\gamma \in \Gamma} A_\gamma$ is closed. 

I have tried various methods but no one worked. I can show that $d(A\cup B) = d(A)\cup d(B）$, where $d(A)$ is the derived set of $A$. Also I know that $d(d(A)) \subset A\cup d(A)$. However, I have no idea about how to use the condition that $d(A_\gamma)$ is closed. Any help will be appreciated. 
 A: Let $A = \cup_{\gamma} A_{\gamma}$. It is clear that $\cup_{\gamma}d(A_{\gamma}) \subset d(A)$. 
We will proceed by showing the complement of $d(A)$ is open.
Suppose that $x \notin d(A)$.  If $x \notin A$ there is an open set $U$ containing $x$ such that $U \cap A = \emptyset$ and we are done.  Hence we will further assume $x \in A$.  We note that $x \notin B$ where 
$$B = \bigcap_{A_{\gamma} \supset \{x\}} d(A_{\gamma})$$
because $x$ is not even in the union of these sets as $x \notin d(A)$.
$B$ is closed by assumption so we can choose an open set $U_1$ containing $x$ such that $U_1 \subset B^{c}$.  $x \notin d(A)$ so we can also choose an open set $U_2$ containing $x$ such that $U_2 \cap A = \{x\}$.
Set $U = U_1 \cap U_2$.  We claim that $U \subset d(A)^c$.
Suppose $y \in U \cap d(A)$.  Note that $y \neq x$. For every $V$ containing $y$ we must have that $U \cap V \cap A = \{x\}$ because $U \subset U_2$ and $y \in d(A)$ (hence $U \cap V \cap A$ contains something other than $y$).  This implies that for any $A_{\gamma}$ containing $x$, $y \in d(A_{\gamma})$ as $\{x\} \subset A_{\gamma} \cap V$ for all open $V$ containing $y$.  In particular, $y \in B$.  Yet $y$ is in $U \subset U_1$, a contradiction.
We are left with $U \cap d(A) = \emptyset$ as desired.
A: I think I found a counter-example: 
For all $n \geq 2$, let $(x^n_k)$ be a sequence in $\left( \frac{1}{n} , \frac{1}{n-1} \right)$ converging to $\frac{1}{n}$. Let $A_n= \{ x^n_k \mid k \geq 0 \} \cup \{ \frac{1}{n} \}$. If $X=[0,+ \infty)$, $d(A_n)= \{ \frac{1}{n} \}$ is closed, but $d \left( \bigcup\limits_{n \geq 2} A_n \right)= \{ \frac{1}{n} \mid n \geq 1 \}$ is not closed in $X$.
