Main Question. Can there be an embedding $j:V\to L$ of the set-theoretic universe $V$ to the constructible universe $L$, if $V\neq L$?

By embedding here, I mean merely a proper class isomorphism from $\langle V,{\in}\rangle$ to its range in $\langle L,{\in}\rangle$, or in other words a quantifier-free-elementary map $j:V\to L$, a class map $j$ for which $x\in y\iff j(x)\in j(y)$.

This embedding concept is considerably weaker than usually considered in set theory, where one typically has embeddings that are at least $\Delta_0$-elementary if not much more. Of course, we may easily refute the existence of nontrivial fully elementary or even of $\Delta_0$-elementary embeddings $j:V\to L$. Those arguments, however, simply fail with this much weaker embedding concept. One can begin to see this by observing that $$j(x)=\{\ j(y)\mid y\in x\ \}\cup\{\ \{0,x\}\ \}$$ defines an embedding $j:L\to L$ with $j(x)\neq x$ for every $x$. In particular, the existence of a nontrivial embedding $j:L\to L$ in this weak sense is consistent with $V=L$ and carries no large cardinal strength, and does not prove the existence of $0^\sharp$.

The question arises in connection with my paper,

where it appears in the final section with the other questions I ask here, among others. I have half an expectation, a gnawing suspicion, however, that this questions may admit an easy answer, and this is why I am asking it here. But I don't know which way the answer will go.

The main theorem of the paper shows that every countable model of set theory $M$ has an embedding $j:M\to L^M$. But the proof establishes the existence of such embeddings only in an external way, using the countability of $M$. The main question above inquires from an internal perspective whether one can ever find such an embedding as a class inside the model.

The existence of such an embedding as a definable class would of course imply $V=HOD$, since one could pull back the canonical order from $L$ to $V$. More generally, if $j$ is merely a class in Gödel-Bernays set theory, then the existence of an embedding $j:V\to L$ implies global choice. So we cannot expect every model of ZFC or of GB to have such embeddings. Can they be added generically? Do they have some large cardinal strength? Are they outright refutable?

There are several more concrete versions of the question.

Question. Does every set $A$ admit an embedding $j:\langle A,{\in}\rangle \to \langle L,{\in}\rangle$? If not, which sets do admit such embeddings?

It follows from the main theorem of the paper that every countable set $A$ embeds into $L$. What about uncountable sets?

Question. Does $\langle V_{\omega+1},{\in}\rangle$ embed into $\langle L,{\in}\rangle$? How about $\langle P(\omega),{\in}\rangle$ or $\langle \text{HC},{\in}\rangle$?

These latter questions are interesting principally when $V$ has non-constructible reals. I would be very interested in learning the answer.

  • 1
    $\begingroup$ Interesting question, although I won't be able to answer. I have a question about the paragraph "The existence of such an embedding ...". Isn't $V=HOD$ undecidable? So your question only makes sense as "for which models $V$ is there an embedding $V \to L$", right? $\endgroup$ – Martin Brandenburg Jul 10 '12 at 7:36
  • $\begingroup$ Martin, indeed $V=HOD$ or $V=L$ are independent of $ZFC$, but the question states that such definable embedding, it is perfectly reasonable that such embedding cannot be defined within the theory and can only be defined externally (or in a stronger theory). $\endgroup$ – Asaf Karagila Jul 10 '12 at 8:14
  • $\begingroup$ I agree that the questions are very interesting, and can be extended if the answer is negative to embeddings into $L[A]$ for some $A$; into $HOD$ or into $HOD[A]$. The first two are very interesting in contexts where $V$ contains sharp for $L$ (or $L[A]$). $\endgroup$ – Asaf Karagila Jul 10 '12 at 8:24
  • $\begingroup$ Martin, yes, since V=HOD can fail, this shows that there are models of ZFC having no definable embedding $j$, and since global choice can fail, there are models of GB having no class embedding $j$. The question, however, is whether there is any model at all with an embedding of the kind I seek. Asaf, yes, indeed, the extension of the question to $L[A]$ and $HOD[A]$ is quite natural. $\endgroup$ – Joel David Hamkins Jul 10 '12 at 10:20
  • $\begingroup$ I recall a theorem of Vopenka that $V$ is always a set-forcing away from $HOD[A]$, so one can ask whether genericity affects embedibility in this context. If the answer is no, then proving things for $HOD[A]$ sort of models should be enough. Right? $\endgroup$ – Asaf Karagila Jul 10 '12 at 18:24

Here are some additional partial results.

Theorem. If there is an embedding $j:V\to L$ in the sense of the question, then for a proper class club of cardinals $\lambda$, we have $(2^\lambda)^V=(\lambda^+)^L$.

Proof. The collection of cardinals $\lambda$ for which $j''\lambda\subset L_\lambda$ is closed and unbounded. For any such $\lambda$, if $A,B\subset\lambda$ are distinct subsets of $\lambda$, then $j(A)\cap L_\lambda\neq j(B)\cap L_\lambda$, and so in $V$ we may place $2^\lambda$ in bijection with the collection of subsets of $L_\lambda$ in $L$, which has size $(\lambda^+)^L$. QED

Corollary. If $0^\sharp$ exists, then there is no embedding $j:V\to L$.

Proof. If $0^\sharp$ exists, then $L$ does not compute such successor cardinals $\lambda^+$ correctly, and so $2^\lambda\geq(\lambda^+)^V\gt(\lambda^+)^L$, which contradicts the previous theorem. QED

Corollary. If the GCH fails on a stationary class of cardinals, then there is no embedding $j:V\to L$.

Proof. If there is an embedding $j:V\to L$ and the GCH fails on a stationary class of cardinals, then there will be a cardinal $\lambda$ at which the GCH fails and for which $j''\lambda\subset L_\lambda$, since these latter cardinals form a class club. But in this case, the theorem shows $2^\lambda=(\lambda^+)^L$, which implies that the GCH holds at $\lambda$, a contradiction. QED

(Update) Improved Corollary. If there is an embedding $j:V\to L$, then the GCH holds everywhere above $\aleph_0$.

Proof. This argument was made by User41953 in a comment below, and the same observation was sent to me from Menachem Magidor by email. The idea is that since we know $0^\sharp$ doesn't exist, we can use the covering lemma. Let $\lambda$ be any uncountable cardinal in $V$. By covering, $j''\lambda \subset B\in L$ with $|B|^V=\lambda$. Since $L$ does not compute successors correctly, we have $|B|^L\lt(\lambda^+)^V$, but by the proof of the theorem $|P^V(\lambda)|≤|P^L(B)|=(\mu^+)^L$, and so $(2^\lambda)^V=(\mu^+)^L\leq (\lambda^+)^V$. QED

The next few observations were made jointly with Menachem Magidor:

Theorem. In the forcing extension $V[G]$ obtained by adding $\omega_1$ many Cohen reals (or more), there is no $j:P(\omega)^{V[G]}\to L$ and indeed no $j:P(\omega)^{V[G]}\to V$.

Proof. Suppose that $V[G]$ is obtained by adding at least $\omega_1$ many Cohen reals, and suppose $j:P(\omega)^{V[G]}\to V$ is an embedding in $V[G]$. Consider $j''\omega$, which is a countable subset of $L$. This set is added by the forcing $V[G]$, and since the forcing is c.c.c., it exists in $V[g]$ for some real $g\in V[G]$. Let $h$ be another Cohen real in $V[G]$ that is mutually generic with $g$. Notice that $j(h)$ is a set in $V$ whose trace on $j''\omega$ is exactly $h$, and so $h\in V[g]$, contrary to mutual genericity. QED

Theorem. For any infinite regular cardinal $\kappa$, in the forcing extension $V[G]$ obtained by adding $\kappa^+$ many subsets to $\kappa$, there is no $j:P(\kappa)^{V[G]}\to V$.

Proof. The same argument works. The set $j''\kappa$ is in $V[g]$ for a single subset $g\subset \kappa$, but then there are other subsets $h\subset\kappa$ with $h\notin V[g]$, although they are determined by $j(h)\in V$ and $j''\kappa\in V[g]$, a contradiction. QED

In other work, currently being written up, Victoria Gitman, Gunter Fuchs and I have proved that if it is consistent that there is a Mahlo cardinal, then it is consistent that there is a transitive model $M$ of ZFC of size and height $\omega_1$, which does not embed into $L_{\omega_1}$.

Meanwhile, the main question seems still to be open, although the evidence is now suggesting that we might hope to refute in ZFC the existence of an embedding $j:V\to L$ when $V\neq L$. Perhaps it would be natural to determined the situation in the forcing extension $L[c]$ obtained by adding a Cohen real.

  • $\begingroup$ Not to be petty about terminology, but aren't the theorems about $0^\#$ and the stationary class where $\sf GCH$ fails are more of corollaries to the first theorem? $\endgroup$ – Asaf Karagila Nov 9 '13 at 13:07
  • $\begingroup$ Yes, that's right. I guess I'll edit, once I'm done with breakfast... $\endgroup$ – Joel David Hamkins Nov 9 '13 at 13:09
  • $\begingroup$ Bon appetit! :-) $\endgroup$ – Asaf Karagila Nov 9 '13 at 13:09
  • $\begingroup$ One approach which might help to solve the case for $L[c]$ where $c$ is Cohen, is to see if $M\subseteq V$ is a model of $\sf ZF$ then $j\colon V\to L$ can be restricted to $M$ as well. In that case, I feel, we may be able to conclude that if $M\subseteq V$ is a model of $\lnot\sf AC$ then there is no such $j$. This will exclude Cohen extensions of $L$, but not Sacks or Miller extensions which are minimal. $\endgroup$ – Asaf Karagila Nov 9 '13 at 13:23
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    $\begingroup$ As a corollary to your corollary - if there is such $j:V\rightarrow L$ then $GCH$ hold everywhere (except maybe at $\aleph_0$). Since $0^\#$ doesn't exist, we can use the covering lemma: let $\lambda$ be uncountable cardinal in $V$. $j^{\prime\prime} \lambda \subset B \in L$ where $|B|^V=\lambda$. Now $|B|^L = \mu < (\lambda^+)^V$ but by the proof of the first theorem, $|P^V (\lambda)| \leq |P^L(B)| = (\mu^+)^L$, so $(2^\lambda)^V = (\mu^+)^L \leq (\lambda^+)^V$. $\endgroup$ – Yair Hayut Nov 10 '13 at 6:41

Assuming $0^\#$ doesn't exist, it's consistent to get a negative answer to those questions:

Assume $(2^{\aleph_0})^V > \aleph_2$ and that ${\aleph_n}^V = {\aleph_n}^L$ for at least $n=1,2,3$.

I claim that there is no embedding $j: P(\omega) \rightarrow L$. Let $A = j^{\prime\prime} \omega$. It is a countable set in $V$ so by the Jensen's covering lemma, there is some $B \in L$ of cardinality at most $\aleph_1$ that covers $A$ (note that since $V$ and $L$ agree on the cardinals, $|B|^L \leq \aleph_1$).

For every $X \in P(\omega)$, the set $j(X)\in L$ codes a subset of $B$: $C_X = \{b \in B | b \in j(X)\} \in L$. Since $j^{\prime\prime} \omega \subset B$, for every $X \neq Y$, $C_X\neq C_Y$. This contradict $|P(B)|^L \leq \aleph_2$.

  • $\begingroup$ But the definition of the power set is not quantifier free; and we only want a quantifier free preservation. Moreover $j''\omega=\omega$, since $k\in\omega$ is an element definable by a quantifier free formula. $\endgroup$ – Asaf Karagila Nov 6 '13 at 21:21
  • $\begingroup$ I answered the second question: is there a embedding from $P(\omega)$ into $L$. The embedding $j$ didn't intend to preserve the power set, but it forced many different sets to be in some $P(B)$. I didn't claim that $j(X)\in P^L(B)$ for some $X \in P^V(\omega)$. I don't agree that $j(k)=k$. For example, in the embedding that Hamkins gave in the question you have $\forall k\,j(k)\neq k$. $\endgroup$ – Yair Hayut Nov 6 '13 at 21:40
  • $\begingroup$ Right. I didn't re-read the entire question again. My bad... $\endgroup$ – Asaf Karagila Nov 6 '13 at 21:52
  • $\begingroup$ Thank you very much for this answer! Your answer got me thinking about the question again, and I posted some new (and old) observations in another answer. It seems that we may be converging to a ZFC resolution of the question. $\endgroup$ – Joel David Hamkins Nov 9 '13 at 12:39

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