# Explicit counter example to Vopěnka's principle in the constructible universe?

Vopěnka's principle is a large cardinal axiom which has many equivalent formulations. One of them, which I find especially appealing, is the following: if the universe is satisfies Vopěnka's principle then no locally presentable category contains a full subcategory which is large (= a proper class) and discrete (= contains no nonidentity morphisms).

On the other hand, I know that many non-set theorists find the constructible universe $V = L$ to be a very appealing place to do mathematics, for a variety of philosophical reasons.

Now, my understanding is that $V =$ a Vopěnka cardinal is a very Large cardinal axiom, and in particular is inconsistent with $V = L$.

So what I would like is an explicit example of a category $C$ which is locally presentable for $V = L$ and a full discrete subcategory of $C$ which is large for $V = L$.

This is very closely related to this previous MO question: Can Vopenka's principle be violated definably?

The difference is that there the OP was asking for a single definable class which violated VP for any universe where VP fails. Here I am asking if such an example can be given in the special case of the constructible universe.

This is a counterexample to Vopenka's principle phrased slightly differently: as "in any proper class of first-order structures, one elementarily embeds into the other."

Working in $V=L$, I claim that $\{L_{\kappa^+}: \kappa\in Card\}$ is a counterexample to Vopenka's principle.

Suppose $\kappa<\lambda$ are cardinals, and $L_{\kappa^+}$ elementarily embeds into $L_{\lambda^+}$, via $j$. Let $\mathcal{U}$ be the set of subsets $X$ of $\kappa$ such that $\kappa\in j(X)$. Clearly $\mathcal{U}$ is an ultrafilter; I claim $\mathcal{U}$ is countably closed. Let $S=\{X_i: i\in\omega\}$ be a sequence of subsets of $\kappa$ such that $X_i\in\mathcal{U}$ for every $i\in\omega$. Note that since $V=L$, the sequence $S$ exists in $L_{\kappa^+}$, and so we can look at $j(S)$. Note first of all that $j(S)$ is an $\omega$-sequence, whose terms are exactly the $j(X_i)$, since $\omega$ can't be moved by $j$. We have $$\forall Y\in j(S), \kappa\in Y,$$ that is, $$\kappa\in\bigcap j(S).$$ But since $S\in L_{\kappa^+}$ and $j$ is elementary, $\bigcap j(S)=j(\bigcap(S))$, so we are done.

So $\mathcal{U}$ is a countably complete ultrafilter on $\kappa$. But this is equivalent to $\kappa$ being measurable, and $V=L$ implies measurable cardinals don't exist.

• Can you explain why $S$ exists in $L_{\kappa^+}$? – Neil Strickland Apr 22 '15 at 10:11
• Also, I guess that the proof is supposed to finish via the argument explained at math.stackexchange.com/a/226616/146477. Is that what you had in mind? – Neil Strickland Apr 22 '15 at 10:13
• $S$ is in $L_{\kappa^+}$ via condensation: take some elementary submodel $M$ of $L_\beta$ (where $S\in L_\beta$ - such a $\beta$ exists since $V=L$) with $L_{\kappa+1}\subseteq M$, $S\in M$, and $\vert M\vert=\kappa$. Now by condensation, the transitive collapse $N$ of $M$ is isomorphic to some $L_\gamma$; by cardinality considerations, we have $\gamma<\kappa^+$. But since $L_\kappa\subset M$, we know that the image of $S$ in $N$ is just $S$ itself - so $S\in L_\gamma$, and hence in $L_{\kappa^+}$. – Noah Schweber Apr 22 '15 at 11:48
• Re: your second comment, yes, the punchline is that measurables can't exist in $L$; I've edited to make that explicit. – Noah Schweber Apr 22 '15 at 12:08
• Meanwhile, it is interesting to note that if Vopenka's principle holds in $V$, when it is larger than $L$, then the class of all $L_{\kappa^+}$ still exists in $V$, and in this case there is an elementary embedding from one $L_{\kappa^+}$ to another. It is just that the embedding does not exist in $L$. So the "counterexample" to VP still exists in $V$, but it is no longer a counterexample there. – Joel David Hamkins Apr 22 '15 at 12:57