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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.

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up vote 8 down vote accepted

To expand what Andres said in the comments above, let's assume that the real $0^\sharp$ exists (see 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.

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Judging by the Just/Weese book, I say that giving $0^\#$ is an example which is difficult to understand in full. The assertion that $0^\#$ exists is quite mind-boggling because it has so many bizarre and counterintuitive consequences (well, countering one's intuition when they have come to the right time to know about $0^\#$ anyway). – Asaf Karagila Dec 22 '12 at 22:16
@AsafKaragila I like to think of $0^\sharp$ as a mouse. It is the simplest mouse beyond $L$, which is either good news for $0^\sharp$ or bad news for mice in general :) – Trevor Wilson Dec 22 '12 at 22:23
Well, I don't know too much about mice. I would like to learn about them but there's just so much to learn before! Either way, you should maybe add a link to somewhere where the description of $0^\#$ is given (and perhaps basic equivalences?). One last thing, I think you're lucky that philosophically most set theorists would agree that $0^\#$ exists. But I bet you'd have problems bringing this to Nelson! :-) – Asaf Karagila Dec 22 '12 at 22:25
Yes, I imagine he wouldn't like it, especially seeing as how $0^\sharp = \left(e^{e^{e^{79}}}\right)^\sharp$. – Trevor Wilson Dec 22 '12 at 22:33
A little remark, since Asaf has mentioned forcing: Under the assumption of the existence of $0^\sharp$, for quite a variety of forcing posets $\mathbb P$ in $L$ (including Cohen forcing), there are (in $V$) many definable sets $G$ that are $\mathbb P$-generic over $L$. – Andrés E. Caicedo Dec 23 '12 at 2:10

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 $\bf L$.

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Though this is correct, there is a strong sense in which we expect $V\ne L$. For example, this is a consequence of large cardinals. And bringing large cardinals into the discussion provides us with plenty of examples: There are reals $x$ that code well-orderings of type $\omega_1^L$ or much larger. The real $0^\sharp$ is not in $L$. Given any measurable cardinal $\kappa$, any witnessing measure $\mu$ is not in $L$, etc. – Andrés E. Caicedo Dec 22 '12 at 21:49
@Andres: True. I will add this. – Asaf Karagila Dec 22 '12 at 21:50
Continuing Andres Caicedo's comment about "a strong sense in which $V\neq L$: If you accept this idea (as most set theorists do), then just imagine flipping a fair coin infinitely often. The resulting binary sequence (representing heads as 0 and tails as 1) is, with probability 1, not in $L$. – Andreas Blass Dec 22 '12 at 22:09
@Andreas: But that would happen in any model where there are only countably many constructible reals. For example if we merely collapsed $\omega_1$, the universe is not going to be that different from $L$, but is still going to have this property which you mentioned. In fact $CH$ even holds in the generic extension. – Asaf Karagila Dec 22 '12 at 22:14
@AndreasBlass Do you have a reference for your comment? – Quinn Culver Dec 23 '12 at 3:24

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