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Stream: deprecated: mathematics

Topic: arbitrary intersections and Set

Matteo Capucci (he/him) (Aug 17 2023 at 08:01):

Recently @Jules Hedges published a blog post where he uses the fact ZFC allows arbitrary intersections of sets. I never made the connection that indeed that's true but in category theory we reject that fact, somehow, and so e.g. coproducts in Set are considered disjoint.

What's going on?

Matteo Capucci (he/him) (Aug 17 2023 at 08:03):

The only answer I can give is: category theorists implicitly use some other foundation where sets/types are disjoint, like... MLTT? But that seems way more constructive than most people would like. Is there something like 'sane ZFC'?

Morgan Rogers (he/him) (Aug 17 2023 at 08:10):

It's not that we assume sets are disjoint, it's that we don't consider intersections and unions of sets to be well-defined unless they are specified with respect to some containing set. Which makes sense, since they would not be invariant under isomorphism otherwise!

Matteo Capucci (he/him) (Aug 17 2023 at 08:17):

right, but then we must be using some other foundations than ZFC, since in ZFC is perfectly fine to

1. have two sets $A$ and $B$ and construct $A \cap B$
2. have two functions $B \leftarrow A \cap B \to A$

Matteo Capucci (he/him) (Aug 17 2023 at 08:17):

(I'm reasoning out loud so I might trip on myself)

Matteo Capucci (he/him) (Aug 17 2023 at 08:20):

now if we form the categorical coproduct $A+B$, that's not disjoint anymore since the terminality of $A \leftarrow \varnothing \rightarrow B$ conflicts with the existence of the span (2)

Morgan Rogers (he/him) (Aug 17 2023 at 08:29):

The categorical coproduct is the disjoint union, not the union ;) Typically it's presented as $\{(a,0) \mid a \in A\} \cup \{(b,1) \mid b \in B\}$.

Matteo Capucci (he/him) (Aug 17 2023 at 08:43):

Morgan Rogers (he/him) said:

The categorical coproduct is the disjoint union, not the union ;) Typically it's presented as $\{(a,0) \mid a \in A\} \cup \{(b,1) \mid b \in B\}$.

but $A \cup B$ has the universal property of the coproduct: for any pair of maps $f,g : A, B \to Z$, there is a unique map $A \cup B \to Z$.. and here's where I trip on myself: there isn't! How would I define $f+g$ on $A\cap B$?

ok thanks for leading me to a resolution :D I feel much better now: Set is safe!

John Baez (Aug 17 2023 at 08:58):

This is a question about the relation between:

• material set theory (the traditional approach, where sets are defined by their members, thus allowing us to talk about the union or intersection of any two sets)

and

• structural set theory (the category-theoretic approach, where it makes sense to talk about the intersection or union of two subsets of a given set, but not of two sets).

What you're calling 'sane ZFC' is usually called structural ZFC.

Matteo Capucci (he/him) (Aug 17 2023 at 09:13):

Yeah but I always thought the two foundations (ETCS+Rep, say, and ZFC) would yield equivalent categories, so I was aghast when it seemed they didn't

Matteo Capucci (he/him) (Aug 17 2023 at 09:13):

In fact they are, as proven e.g. by Shulman:
image.png

John Baez (Aug 17 2023 at 09:19):

Okay, I didn't understand what you were saying ("now if we form the categorical coproduct A+B, that's not disjoint anymore"), so I didn't get that you thought you were getting nonequivalent categories.

Matteo Capucci (he/him) (Aug 17 2023 at 22:01):

I was confused by myself so it's more than fair if I also confused you :sweat_smile:

John Baez (Aug 18 2023 at 15:34):

Confusion is contagious. :upside_down: