Let $\kappa$ be a cardinal, let $\mathbf{H}_\kappa$ be the set of hereditarily $\kappa$-small sets, and let $\mathbf{Set}_{< \kappa}$ be the full subcategory of $\mathbf{Set}$ corresponding to $\mathbf{H}_\kappa$. I am interested in properties of $\kappa$ that are expressible in terms of $\mathbf{Set}_{< \kappa}$ (regarded as an abstract category, up to equivalence). For example:

  • $\kappa = 0$ if and only if $\mathbf{Set}_{< \kappa} = \emptyset$.
  • $\kappa = 1$ if and only if $\mathbf{Set}_{< \kappa}$ is equivalent to the terminal category.
  • $\kappa$ is either $1$ or infinite if and only if $\mathbf{Set}_{< \kappa}$ has finite limits.
  • $\kappa$ is either $1$ or a strong limit cardinal if and only if $\mathbf{Set}_{< \kappa}$ is cartesian closed.

Question. Is the property "$\kappa$ is regular" of this type?

Answer. Yes. We assume $\kappa$ is infinite, so that $\mathbf{Set}_{< \kappa}$ has finite limits. For each object $X$ in $\mathbf{Set}_{< \kappa}$, let $\Gamma (X)$ be the set of morphisms $1 \to X$ in $\mathbf{Set}_{< \kappa}$, where $1$ is a fixed terminal object in $\mathbf{Set}_{< \kappa}$. Pullback then defines a functor $$(\mathbf{Set}_{< \kappa})_{/ X} \to \prod_{x \in \Gamma (X)} \mathbf{Set}_{< \kappa}$$ and $\kappa$ is regular if and only if this functor is (fully faithful and) essentially surjective on objects for every object $X$ in $\mathbf{Set}_{< \kappa}$.

However, I find the above answer unsatisfactory, for two main reasons:

  • It relies on the axiom of replacement – which is perhaps unavoidable because of the nature of the definition of "regular cardinal".
  • It also exploits the fact that every set is the disjoint union of its elements.

Is there a better answer?

  • $\begingroup$ Well, $\aleph_0$ is strongly inaccessible (in the property of the subcategory being cartesian closed). $\endgroup$ – Asaf Karagila Dec 9 '15 at 16:06
  • $\begingroup$ That is my convention too, but some people like to think that strongly inaccessible cardinals are uncountable. $\endgroup$ – Zhen Lin Dec 9 '15 at 16:30
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    $\begingroup$ $\kappa$ does not have to be inaccessible for $\mathbf{Set}_{< \kappa}$ to be cartesian closed. Indeed, you just need cardinalities $<\kappa$ to be closed under exponentiation, which is true iff $\kappa$ is a strong limit cardinal. $\endgroup$ – Eric Wofsey Jun 28 '16 at 5:43

I know this is very old by now, but couldn't you say that $\kappa$ is regular iff $\mathbf{Set}_{<\kappa}$ is cocomplete? That is, for every $X\in\text{Ob}(\mathbf{Sets}_{<\kappa})$ and every diagram (functor) $D:X\to\mathbf{Set}_{<\kappa}$, the colimit $\text{colim }D$ exists in $\mathbf{Set}_{<\kappa}$? Then you're not talking about elements at least, and the definition doesn't use Replacement to work (but can be seen as postulating the Replacement axiom relativised to $\textbf{Set}_{<\kappa}$).

It seems like it then holds that $\kappa$ is a strong limit iff $\textbf{Set}_{<\kappa}$ is an elementary topos and that $\kappa$ is strongly inaccessible iff $\textbf{Set}_{<\kappa}$ is a Grothendieck topos, which is quite satisfying.

  • $\begingroup$ Here I take a class $X$ to be small iff $X\in\textbf{H}_\kappa$, in the definition of cocompleteness. $\endgroup$ – Dan Saattrup Nielsen Nov 10 '16 at 10:32

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