I'm working through the the "Families" chapter in "Naive Set Theory" by Paul Halmos, and found this note in the final exercise:
Prove also (with appropriate provisos about empty families) that $\bigcap_i X_i \subset X_j \subset \bigcup_i X_i$ for each index $j$ and that intersection and the union can in fact characterized as the
extreme solution of these inclusions [emphasis mine]. This means that if $X_j \subset Y$ for each index $j$, then $\bigcup_i X_i \subset Y$, and that $\bigcup_i X_i$ is the only set satisfying this minimality condition; the formulation for intersections is similar.
What does it mean to say that intersection and union are the "extreme solution of these inclusions"? The empty set $\emptyset$, for example, seems like a better candidate for the extreme solution on the left side of the inclusion relationship, and any set disjoint from any $X_j$ unioned in with the full union would increase the bounds.
Does it perhaps mean that these are the tightest possible bounds?
This exercise was also covered in Exercise in Halmos set theory book., but not this aspect.
Best Answer
I agree that the note isn't very clearly stated. What Halmos means is that $\bigcup_i X_i$ contains every $X_j$ and is contained in any other set $Y$ that contains every $X_j$. I.e., $\bigcup_i X_i$ is minimal amongst all sets $Y$ that contain every $X_j$.