Easing notation on unions and intersections. On writing some notes on Topology, I came across proving that the topology generated by any collection is the set of unions of finite intersections of elements of the given collection. In trying to be concise, I argued (essentially, not ipsis litteris) as follows:

Let $\Omega$ be the set of unions of finite intersections of elements of the collection $\Phi$, and let $\tau_m$ be the topology generated by $\Phi$ (the smallest topology containing $\Phi$). From the axioms of topology, it follows that $\Omega \subset \tau_m$. It therefore suffices to prove that $\Omega$ is a topology. But this follows from the equalities
$$\bigcup_{\alpha \in A}  \left( \bigcup_{\beta \in B_{\alpha}} \left( \bigcap_{i \in I_{n(\alpha, \beta)}} U_{\alpha,\beta,i}\right) \right) = \bigcup_{\beta \in \bigsqcup_{\alpha \in A} B_{\alpha}} \left( \bigcap_{i \in I_{n(\alpha, \beta)}} U_{\alpha,\beta,i} \right)  ,$$
$$ \left( \bigcup_{\alpha \in A} \left( \bigcap_{i \in I_{n(\alpha)}} U_{\alpha,i}\right) \right) \cap \left( \bigcup_{\beta \in B} \left( \bigcap_{j \in I_{n(\beta)}} U_{\beta,j}\right) \right) = \bigcup_{(\alpha,\beta) \in A \times B} \left( \bigcap_{\xi \in  \bigsqcup_{\omega \in \{\alpha,\beta \} } I_{n(\omega)}} U_{\xi} \right).$$

I think that my notation got overloaded. Does anyone have a suggestion on cleaner notation for this?
 A: I think in part many of these issues could be cleared up by adopting some kind of convention regarding bracketing. For example, the left-hand side of your first equation could just as well be written as
$$\bigcup_{\alpha \in A} \bigcup_{\beta \in B_{\alpha}} \bigcap_{i \in I_{n(\alpha,\beta)}} U_{\alpha,\beta,i}$$
This convention is standard, and eliminates all but two pairs of parentheses in your quote, namely outer parentheses in the terms either side of the $\cap$ symbol on the left-hand side of your second equation.
You could also spell out the equations in terms of quantifiers. Indeed, given sets $X$ and $Y$, $X=Y$ if and only if $x \in X \Leftrightarrow x \in Y$ for all possible values of $x$; then unions can be written in terms of existential quantifiers ($\exists$) and intersections in terms of universal quantifiers ($\forall$).
But frankly, if your goal is to communicate mathematics, then it might be worth going into more detail in your notes, at the expense of conciseness.
A: It might be useful to do it in two steps, using the notion of a base for the topology. To recap, for a topological space $(X,\mathcal{T})$, a base for $X$ is a collection $\mathcal{B} \subseteq \mathcal{T}$ such that for every open set $O$, there exists some $\mathcal{B'} \subseteq \mathcal{B}$ such that $\cup \mathcal{B'} = O$. So every open set is a union of sets from the base, and all base sets are themselves open.
Now it is well-known that if we have a set $X$, without a topology, and we have a collection $\mathcal{B} \subseteq \mathcal{P}(X)$, then $\mathcal{B}$ is a base for a (uniquely determined) topology $\mathcal{T}$ on $X$ iff it satisfies two conditions:


*

*$\cup \mathcal{B} = X$.

*For all $B_1, B_2 \in \mathcal{B}$ and for all $x \in B_1 \cap B_2$, there exists some $B_3 \in \mathcal{B}$ such that $x \in B_3 \subseteq B_1 \cap B_2$.


and this topology is defined by $\mathcal{T} = \{ \cup \mathcal{B'}: \mathcal{B'} \subseteq \mathcal{B}\}$, i.e. all unions of subfamilies of the base. In fact, $\mathcal{T}$ is clearly the smallest topology that contains $\mathcal{B}$ as a subset (because topologies are closed under unions, so these unions have to be in there anyway; the second of the conditions is in fact used to show that finite intersections of such unions are in fact already in it as well). This is well-covered in Munkres or any standard topology text book.
Now, if we have any collection $\mathcal{S}$ of subsets of $X$ (a set again), then define $\mathcal{B} = \{ \cap \mathcal{S'}: \mathcal{S'} \subseteq \mathcal{S} , \mathcal{S'} \text{ finite } \}$, i.e. all finite intersections from $\mathcal{S}$, where $\cap \emptyset = X$ (by convention / logic).
One checks that this $\mathcal{B}$ satisfies the conditions mentioned above for being a base: the empty intersection gives the first, and our newly defined $\mathcal{B}$ is closed under intersections, which trivially gives the second condition, as we can always take $B_3 = B_1 \cap B_2$. 
But looking at the theorem above we see that all unions from $\mathcal{B}$ exactly generate the topology generated by $\mathcal{B}$. So If we look at any topology $\mathcal{T'}$ that contains $\mathcal{S}$, then $\mathcal{B} \subseteq \mathcal{T'}$ as topologies are closed under finite intersections and so $\mathcal{T}$ (the topology of unions of $\mathcal{B}$) $\subseteq \mathcal{T'}$ as well. So indeed the topology generated by $\mathcal{S}$ is exactly the set of all unions of finite intersections from $\mathcal{S}$, as claimed. 
Setting it up this way, we introduce the useful notion of a base (which you need anyway) and avoids the computational approach in your proof.
