Example of a set that is in $\mathbf V$ but not in $\mathbf L$ Let $\mathbf V$ denote the cumulative hierarchy and let $\mathbf L$ denote Gödel's constructible universe. We then have $\mathbf L \subseteq \mathbf V$.
Would someone give me an example of a set that is in $\mathbf V \setminus \mathbf L$? Many thanks for your help.
 A: To expand what Andres said in the comments above, let's assume that the real $0^\sharp$ exists (see http://en.wikipedia.org/wiki/Zero_sharp.)  This follows from sufficiently strong large cardinal axioms, such as the existence of a measurable cardinal.  Then $0^\sharp$ is not in $L$.  One thing that makes this example special (e.g. compared to a real that is Cohen-generic over $L$) is that $0^\sharp$ is definable via a definition that is absolute to any transitive model of set theory that contains it and contains all the countable ordinals.  In particular, we have $(0^\sharp)^{L[0^\sharp]} = 0^\sharp$ and for any forcing extension of $V$ by a generic filter $G$ we have $(0^\sharp)^{V[G]} = 0^\sharp$. This is a similar kind of absoluteness to that which $L$ itself has.  So I think it's fair to think of the statement "$0^\sharp$ exists" in philosophical terms as asserting the existence of a "definite object" that is not in $L$.
EDIT:  The reason I think it is appropriate to get philosophical here is the the question asks for an "example of a set" not in $L$.  This does not quite make sense formally.  One could formalize this as asking for an example of a formula that defines a set not in $L$, which I think is more or less what Asaf did, or one could simply not formalize it and pretend that a set is an object that a set theorist can can take out of his or her pocket to show people, which I think is a more attractive notion.
A: No. It's not possible to give such example. Not explicitly, not in the generality implied by the question, anyway.
The reason is that it is consistent that $\bf V=L$. In such case it is not possible to give a counterexample.
However suppose that $M$ is a countable transitive model of ZFC+$\bf V=L$. We can extend $M$ by forcing and add new sets. Let $N=M[G]$ a generic extension of $M$, then $N$ is also a countable transitive model. Therefore $M=\mathbf L^M=\mathbf L^N$. But $N$ has new sets which are not in $M$, and therefore these sets satisfy the required property.
Now consider working internally in $N$, then $N=\bf V$ now, and $M=\bf L$. So we have some generic set $G\in N\setminus M=\bf V\setminus L$. However describing in explicit details such set would be impossible for the same reason it is impossible to describe a well-order of $\mathbb R$ without using the axiom of choice. It is simply consistent that there is none, unless we assume this is not the case.
As Andres Caicedo comments, there are plenty of sets that are not in $\bf L$ but their existence requires us to assume additional axioms. For example if $\kappa$ is a measurable cardinal and $\cal U$ is a $\kappa$-complete ultrafilter on $\kappa$, then $\cal U\notin\bf L$.
There are also axioms that when assumed assure that there are sets not in $\bf L$ which do not require an additional consistency strength. For example if we assume that $CH$ fails then $2^{\aleph_0}>\aleph_1$, and then we must have that almost all the real numbers are not in $\bf L$.
