5
$\begingroup$

A category $\mathcal{C}$ is called $\textbf{discrete}$ if the only morphisms are identity morphisms.

Consider the following weaker notion: a category $\mathcal{C}$ is called $\textbf{totally disconnected}$ if $\text{Hom}_\mathcal{C}(C,D)=\varnothing$ for all $C\neq D$.

Vopenka's principle ($\textbf{VP}$) states that a full large subcategory $\mathcal{D}$ of a locally presentable category $\mathcal{C}$ cannot be discrete. I want to consider an anaologus statement:

($\textbf{VP2}$) Full large subcategory $\mathcal{D}$ of a locally presentable category $\mathcal{C}$ cannot be totally disconnected.

Since discrete categories are totally disconnected, $\textbf{VP2}$ implies $\textbf{VP}$. Hence $\textbf{VP2}$ can't be proven in $\textbf{ZFC}$, because $\textbf{VP}$ can't. Now I would like to know:

(1) Is there a counter-example to $\textbf{VP2}$ in $\textbf{ZFC}$? As far as I know, no counter-example to $\textbf{VP}$ has been found but since $\textbf{VP2}$ is in principle a stronger statement, perhaps we can construct one here?

(2) If not, what is the relation between $\textbf{VP}$ and $\textbf{VP2}$ inside $\textbf{ZFC}$? Are they equivalent? Or does at least consistency of $\textbf{VP}$ implies that of $\textbf{VP2}$?

(I wrote $\textbf{ZFC}$ as my set theory of choice but if there are any problems in using such a weak theory, feel free to consider some other).

Thanks!

Improve this question
$\endgroup$
3
  • $\begingroup$ The category of fields is a disconnected full large subcategory of the category or rings, which is locally finitely presentable. Or am I missing something? $\endgroup$
    – Alex Kruckman
    Commented Mar 13, 2022 at 13:14
  • $\begingroup$ @AlexKruckman The category of fields is "not connected" but is not "disconnected" in the sense I defined above. I guess better term for my "disconnected" would be "totally disconnected". I will edit my question appropriately. $\endgroup$
    – T.Ch.
    Commented Mar 13, 2022 at 13:33
  • 1
    $\begingroup$ Ah, I see! I've edited to make the quantifier on $C$ and $D$ more explicit. $\endgroup$
    – Alex Kruckman
    Commented Mar 13, 2022 at 14:29

1 Answer 1

Reset to default
6
$\begingroup$

VP2 is equivalent to VP because every set carries a rigid binary relation. This is similar as Lemma 6.3 in my book with Adámek. In fact, VP2 is an original formulation of VP (see Jech, Set Theory).

Improve this answer
$\endgroup$
1
  • $\begingroup$ Thank you very much! $\endgroup$
    – T.Ch.
    Commented Mar 17, 2022 at 13:04

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .