problem 1.6 in open-logic-complete

[rishihho]

it might be helpful to mention that B is a set of sets because if there's an element in B that isn't a set, the union on all of B wouldn't be well defined.

[rzach]

I don't think that's necessary. \bigcup B is the set of all elements of elements of B. It doesn't matter if B happens to have an element that's not a set; it just isn't an element of \bigcup B. If B contains no sets at all, e.g., then \bigcup B = \emptyset. To give the proof, knowing that A \in B is all that's needed.

[rishihho]

just to clarify with an example, if B = {{1, 2}, 3, {4}}, then \bigcup B = {1, 2, 4}?