Irreducible set Let $X$ be a topological space
Let $A$ be the set of all closed, irreducible subsets of $X$ equipped with a topology that contains all sets of the form $V(U)=\{a\in A| a\cap U\neq\emptyset, \text{where $U\in$ (the topology of $X$, $T_X$)}\}$. So the topology of $A$ is $\{V(U)|U\in T_X\}$
What then is a closed, irreducible subset of $A$? 
Sets of sets confuse me. If anyone has any good way of thinking about them do please divulge!
 A: I'm going to change the notation a bit to something with which I'm more comfortable. Let $\langle X,\tau\rangle$ be a topological space, and let $\mathscr{A}$ be the set of irreducible closed subsets of $X$. For $U\in\tau$ let $$\mathscr{V}(U)=\{A\in\mathscr{A}:A\cap U\ne\varnothing\}\;.$$ 
It's easy to check that $\mathscr{V}(U)\cap\mathscr{V}(W)=\mathscr{V}(U\cap W)$ for any $U,W\in\tau$ and that $$\mathscr{V}\left(\bigcup\mathscr{U}\right)=\bigcup\{\mathscr{V}(U):U\in\mathscr{U}\}$$ for any $\mathscr{U}\subseteq\tau$, so $\mathfrak{T}=\{\mathscr{V}(U):U\in\tau\}$ is a topology on $\mathscr{A}$.
Now let $\mathscr{F}\subseteq\mathscr{A}$, and suppose that $\mathscr{F}\cap\mathscr{V}(U)\ne\varnothing\ne\mathscr{F}\cap\mathscr{V}(W)$ for some $U,W\in\tau$. If $\mathscr{F}$ is irreducible, there must be some $A\in\mathscr{F}\cap\mathscr{V}(U)\cap\mathscr{V}(W)\ne\varnothing$; then $U\cap A\ne\varnothing\ne W\cap A$, so $A\cap U\cap W\ne\varnothing$, and in particular $U\cap W\ne\varnothing$ and $A\in\mathscr{F}\cap\mathscr{V}(U\cap W)$. Conversely, if $U,W\in\tau$ and $\mathscr{F}\cap\mathscr{V}(U\cap W)\ne\varnothing$, then clearly $\mathscr{F}\cap\mathscr{V}(U)\ne\varnothing\ne\mathscr{F}\cap\mathscr{V}(W)$. Thus, for irreducible $\mathscr{F}\subseteq\mathscr{A}$ and $\mathscr{V}(U),\mathscr{V}(W)\in\mathfrak{T}$ we have $$\mathscr{F}\cap\mathscr{V}(U)\ne\varnothing\ne\mathscr{F}\cap\mathscr{V}(W)\;\text{ iff }\;\mathscr{F}\cap\mathscr{V}(U\cap W)\ne\varnothing\;.$$
Now $\mathscr{F}\cap\mathscr{V}(U)\ne\varnothing\ne\mathscr{F}\cap\mathscr{V}(W)$ iff there are $A,B\in\mathscr{F}$ such that $A\cap U\ne\varnothing\ne B\cap W$, so for irreducible $\mathscr{F}$ we must have $$\exists A,B\in\mathscr{F}(A\cap U\ne\varnothing\ne B\cap W)\;\text{ iff }\;\exists A\in\mathscr{F}(A\cap U\cap W\ne\varnothing)\;.$$ This can be stated more simply: $$U\cap\bigcup\mathscr{F}\ne\varnothing\ne W\cap\bigcup\mathscr{F}\;\text{ iff }\;\exists A\in\mathscr{F}(A\cap U\cap W\ne\varnothing)\;.\tag{1}$$
Let $F=\bigcup\mathscr{F}$. Then $\exists A\in\mathscr{F}(A\cap U\cap W\ne\varnothing)$ iff $F\cap U\cap W\ne\varnothing$ (since every $A\in\mathscr{F}$ is irreducible in $X$), so $(1)$ can be simplified further to $$U\cap F\ne\varnothing\ne W\cap F\;\text{ iff }\;(U\cap W)\cap F\ne\varnothing\;,$$ which simply says that $F$ is irreducible in $X$.
In other words, $\mathscr{F}\subseteq\mathscr{A}$ is irreducible in $\mathscr{A}$ iff $\bigcup\mathscr{F}$ is irreducible in $X$.
