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Suppose I have a partially ordered index set $(I, \geq_b)$, which indexes a family of sets $\{S_i\}_{i \in I}$.

If we let $i \in I$, is it permissible to perform a generalized intersection as follows? $$ \bigcap_{j \geq_bi} S_j $$And, also for a generalized union?

Arturo Magidin
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cisprague
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    You can form the intersection or union of any family of sets. – Brian M. Scott Mar 17 '21 at 19:01
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    @BrianM.Scott Not quite true. You can't take the intersection of an empty family of sets. – Mark Saving Mar 17 '21 at 19:02
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    @MarkSaving Sure you can! (Oh wait - are you not working in $\mathsf{NFU}$? :P) – Noah Schweber Mar 17 '21 at 19:03
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    @MarkSaving: I’m aware of that. In this case I thought it more important to make the basic point. – Brian M. Scott Mar 17 '21 at 19:05
  • @MarkSaving isn't the intersection of an empty family of sets just the universe, i.e., if the universe is real numbers then $\bigcap_{i \in \emptyset} S_i = \mathbb{R}$. – cisprague Mar 17 '21 at 20:20
  • @cisprague If you have defined a universe, this is correct. But in general, the "universe" is "all sets", and thus the intersection would be the "set of all sets", which does not exist. – Mark Saving Mar 17 '21 at 20:21
  • @MarkSaving, so if I do happen to encounter $\bigcap_{i \in \emptyset} S_i$ while working on a particular paper, how should I define that $\bigcap_{i \in \emptyset} S_i = \mathbb{R}$? – cisprague Mar 17 '21 at 20:23

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Yes. This is just another case of an intersection indexed by a set - here, fixing $i\in I$ the indexing set is $$J=\{j\in I: j\ge_bi\}$$ so that we're looking at $\bigcap_{j\in J}S_j$ or $\bigcup_{j\in J}S_j$ respectively. This really has nothing to do with the partial order.

I suspect the point at issue is that we initially think of unions/intersections as happening in some order. While this framework is satisfactory at first, once we move into more advanced topics it's really essential to forget the idea of ordering our sets entirely and just think about intersections/unions of arbitrary families of sets. (That said, note that things are a bit weird if we try to take the intersection or union of no sets, but this is an edge case you can ignore at first.)

EDIT: to address the comment below, yes, this is fine. We often abbreviate something like $$\bigcup_{j\in\{x: \varphi(x)\}}A_j$$ with $$\bigcup_{\varphi(j)}A_j,$$ so e.g. $$\bigcup_{j\ge_bi}S_j=\bigcup_{j\in\{x: x\ge_bi\}}S_j.$$ This is completely unambiguous and quite common; it's analogous to something like $$\sum_{1\le i<j\le n}{i\choose j}.$$

Noah Schweber
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    Which is why I routinely describe the base generated by a subbase $\mathscr{S}$ as $$\left{\bigcap\mathscr{F}:\mathscr{F}\subseteq\mathscr{S}\text{ is finite}\right}$$ rather than as $$\left{\bigcap_{k=1}^nS_k:n\in\Bbb Z^+\text{ and }S_1,\ldots,S_n\in\mathscr{S}\right}$$ or the like. The sooner people get used to thinking in those terms, the better. – Brian M. Scott Mar 17 '21 at 19:07
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    @BrianM.Scott I thoroughly agree! – Noah Schweber Mar 17 '21 at 19:36
  • Sorry, my question was more about notation. In my case it is much more convenient to write $\bigcap_{j \geq_b i} S_i$. Is such a notation permissible? – cisprague Mar 17 '21 at 20:26
  • @cisprague Yes, that's fine. – Noah Schweber Mar 17 '21 at 21:17