# Does this notation for arbitrary unions and intersections make sense?

Suppose I have a set $A$ containing sets of points. If $A$ can be infinite, I've read that the union of the elements of $A$ can be written as

$\underset{i\in I}\bigcup A_i$

where $I$ is an index set.

However, I've also seen the following notation, which seems nice in that you don't have to define an index set, even to address infinite sets:

$\underset{B\in A}\bigcup B$

1. Are there any downsides to using this notation?
2. Using this notation, is it OK to define a new set based on each $B$ we "pick out" of $A$? E.g. $\underset{B\in A}\bigcup \{x\in B : x > 0\}$.
• These notations are all fine, so you can use whichever is most convenient. Feb 17, 2017 at 16:25

The two notations are equivalent. Another way of writing $\bigcup_{B \in A} B$ is just as $\bigcup A$, which implies you are taking the union of all the elements of $A$. In fact, this is basically the axiom of union in ZFC.
If you have some index set $I$, that means you also must have some bijection $f: I \to A$ (otherwise it's not really an index set). We can then simply apply replacement to create a set $B$ such that $i \in I \iff f(i) \in B$, so $B$ is simply the range of $f$. Then $\bigcup B$ is the union of all sets indexed by $I$.
Taking the union while also restricting which elements you include in the union set is also fine, as this is basically an application of union followed by comprehension. In other words $\underset{B\in A}\bigcup \{x\in B : x > 0\}$ is the same as first taking $C = \bigcup A$ then $D = \{c \in C : c > 0\}$ (the reverse also works where you first apply the restriction and then take the union).
I don't know any downside of using the notation $\bigcup_{B \in A}B$ but the first one whith the Index set gives you some additional posibilities. For instance $I$ could be a directed set (or an ordered set if you prefer) and in that case you could write $\bigcup_{i\geq i_0}B_i$ for some $i_0 \in I$, which might be useful if you're dealing with nets (or sequences if you let $I=\mathbb{N}$) of sets.