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1. removing query boxes

+– {: .query} Madeleine Birchfield: Wouldn’t a cardinal number be an object of the decategorification of the category Set, just as a natural number is an object of the decategorification of the category FinSet? =–

+– {: .query} Roger Witte First of all sorry if I am posting in the wrong place

While thinking about graphs, I wanted to define them as subobjects of naive cardinal 2 and this got me thinking about the behaviour of the full subcategories of Set defined by isomorphism classes. These categories turned out to be more interesting than I had expected.

If the background set theory is ZFC or similar, these are all large but locally small categories with all hom sets being isomorphic. They all contain the same number of objects (except 0, which contains one object and no non-identity morphisms) and are equinumerous with Set. Each hom Set contains $N^N$ arrows. In the finite case $N!$ of the morphisms in a particular hom set are isomorphisms. In particular, only 0 and 1 are groupoids. I haven’t worked out how this extends to infinite cardinalities, yet.

If the background theory is NF, then they are set and 1 is smaller than Set. I haven’t yet worked out how 2 compares to 1. I need to brush up on my NF to see how NF and category theory interact.

I am acutely aware that NF/NFU is regarded as career suicide by proffesional mathematicians, but, fortunately, I am a proffesional transport planner, not a mathematician.

Toby: Each of these categories is equivalent (but not isomorphic, except for 0) to a category with exactly one object, which may be thought of as a monoid. Given a cardinal $N$, if you pick a set $X$ with $N$ elements, then this is (up to equivalence, again) the monoid of functions from $X$ to itself. The invertible elements of this monoid form the symmetric group, with order $N!$ as you noticed. Even for infinite cardinalities, we can say $N^N$ and $N!$, where we define these numbers to be the cardinalities of the sets of functions (or invertible functions) from a set of cardinality $N$ to itself.

From a structural perspective, there's no essential difference between equivalent categories, so the fact that these categories (except for 0) are equinumerous with all of Set is irrelevant; what matters is not the number of objects but the number of isomorphism classes of objects (and similarly for morhpisms). That doesn't mean that your result that they are equinumerous with Set is meaningless, of course; it just means that it says more about how sets are represented in ZFC than about sets themselves. So it should be no surprise if it comes out differently in NF or NFU, but I'm afraid that I don't know enough about NF to say whether they do or not.

By the way, every time you edit this page, you wreck the links to external web pages (down towards the bottom in the last query box). It seems as if something in your editor is removing URLs. =–

Anonymous

2. adding section about cardinals in homotopy type theory and a reference from the HoTT book

Anonymous

• CommentRowNumber3.
• CommentAuthorjkabrg
• CommentTimeOct 31st 2022
• (edited Oct 31st 2022)
I don't agree with the following bit:

''The usual way to define an ordering on cardinal numbers is that |S1|≤|S2| if there exists an injection from S1 to S2:

(|S1|≤|S2|):⇔(∃(S1↪S2)).
Classically, this is almost equivalent to the existence of a surjection S2→S1, except when S1 is empty''

I believe this isn't true? Unless by ''classically'', you're assuming Choice. The analogue of Cantor-Schroeder-Bernstein where injections are replaced with surjections, is equivalent to the Axiom of Choice. I believe that without Choice, the cardinal numbers under the above ordering are not well-ordered, or even linearly ordered, but merely partially ordered.
• CommentRowNumber4.
• CommentAuthorDavidRoberts
• CommentTimeOct 31st 2022
• (edited Oct 31st 2022)

I think the “this” is in the phrase

Classically, this is almost equivalent to the existence of a surjection

is not referring to the equivalence, but to the existence of an injection. So you have, classically, $(\exists\, S_1\hookrightarrow S_2) \Leftrightarrow (S_1\neq \emptyset \Rightarrow \exists\, S_2 \twoheadrightarrow S_1)$.