TITLE: Example of a set that is in $\mathbf V$ but not in $\mathbf L$
QUESTION [7 upvotes]: 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.

REPLY [8 votes]: 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.