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. 
 A: 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.
