Q(1): Can the category of partial orders be fully embedded in the category of linear orders?

Vopěnka's principle, or VP, is a very intriguing axiom with many equivalent forms and consequences spanning universal logic (in the Barwise sense), large cardinals, model theory, and category theory.

VP for $C$ for a (large locally presentable) category $C$ means that there is no large discrete full subcategory of $C$. It is widely known that VP for $C$ for most "sufficiently expressive" categories $C$ is simply equivalent to all of VP (see Joel David Hamkin's answer to this question).

The notion of "sufficiently expressive" can be formalized as follows: if $A$ fully embeds into $B$ and VP for $B$ holds, then any large discrete full subcategory of $A$ necessarily produces large discrete full subcategory of $B$. So, perhaps a suitable notion of "sufficiently expressive" is simply a full embedding. For example, the categories of structures of a given language can be fully embedded in the category of graphs, showing that, for example, VP for graphs is the same as VP for structures.

This leads me to the question above; an affirmative answer (which I believe is impossible) would also definitively show that VP for linear orders is indeed equivalent to VP itself. However, I have strong doubts about this. Perhaps it needs to be weakened?

Q(2): Is the category of linear orders a reflective subcategory of the category of partial orders?

It isn't immediately obvious that this even relates to VP anymore. However, as shown in Adamek-Rosicky's Locally Presentable and Accessible Categories the following is equivalent to WVP (Weak Vopenka's principle, weakened form):

For $C$ a locally presentable category, every full subcategory $D↪C$ which is closed under limits is a reflective subcategory.

Now it becomes clear! The category of partial orders is certainly locally presentable, and the category of linear orders is closed under limits. Under the Weak Vopenka's principle therefore, the answer to Q(2) is affirmative! But is using such a powerful axiom unnecessary? The original hypothesis also implies Q(2); if strong large cardinal axioms are indeed necessary for Q(2), that doesn't bode very well for our friend Q(1).

This leads me to Q(3), where we investigate the consistency strengths of Q(1) and Q(2):

Q(3): Is it consistent that Q(2) fails? If so, what is the consistency strength of Q(2)? What is the consistency strength of Q(1)?

And finally, of course, the last question that motivated this whole thing in the first place:

Q(4): Is VP for linear orders equivalent to VP?

Adámek, Jiří; Rosický, Jiří, Locally presentable and accessible categories, London Mathematical Society Lecture Note Series. 189. Cambridge: Cambridge University Press. xiv, 316 p. (1994). ZBL0795.18007.

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    $\begingroup$ The category of linear orders is not closed under limits (or even products) in the category of partial orders. $\endgroup$ Dec 8, 2019 at 10:19
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    $\begingroup$ Ah! Are your morphisms $<$-preserving functions, rather than $\leq $-preserving functions? $\endgroup$ Dec 8, 2019 at 10:30
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    $\begingroup$ Your title and your first sentence are different. $\endgroup$ Dec 8, 2019 at 10:32
  • $\begingroup$ In any case, the category of linear orders is not a reflective subcategory of the category of partial orders: where could the poset with two incomparable elements be sent? $\endgroup$ Dec 8, 2019 at 11:08
  • $\begingroup$ @JeremyRickard Oh yes! Nice catch haha $\endgroup$ Dec 8, 2019 at 18:16

1 Answer 1


Expanding on Jeremy's comments,

Q1) No, the category of partial orders does not admit a faithful functor $F$ into the category of linear orders. The poset $\{l,r\}$ of two incomparable elements has a nonidentity automorphism $n$. If the global elements $l$, $r$ satisfy $F(l) < F(r)$ or $F(r) < F(l)$, then $F(n)$ is not monotone, so $F(l) = F(r)$ and $F$ is not faithful.

Q2) Neither the category of linear orders with $\leq$- or $<$ preserving functions is closed under products, and therefore cannot be a reflective subcategory. For $\mathbf{Lin}_\leq$, by considering global elements, we must have that $2 \times 2 \cong 4$, and then the function $(\pi_2, \pi_1)$ isn't monotone. For $\mathbf{Lin}_<$, there is no terminal object.


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