Suppose $V$ is a model of ZF. Within $V$ we have $L$ which is a model of ZFC, furthermore $L[A]$ is a model of choice for every $A\in V$.

Suppose $A=\emptyset$ then clearly $L[A]=L$, furthermore if $A\in L$ then $A\cap L\in L$, therefore $L[A]=L$. Recall also that if $A' = L[A]\cap A$ then $L[A] = L[A']$.

On the other extreme, suppose $A$ is an amorphous set (an infinite set that every subset of it is either finite or co-finite). Consider $A'=A\cap L[A]$, we have that $A'\in L[A]$ which is a model of choice, so $A'$ cannot be infinite - since infinite subsets of amorphous sets are themselves amorphous. Therefore $A'$ is finite, despite not being able to prove that (at the moment anyway) I have a strong intuition that $A'=A\cap L$ and therefore $L[A]=L$.

(this conjecture stems from noticing that amorphous sets are, as the name suggests, amorphous. There is no actual reason that any element would be "preferred" into $L[A]$ over another, unless it was already in $L$. Since there can only be finitely many constructible elements in an amorphous set this somewhat supports my intuition)

Is this conjecture about amorphous sets true? Can it be extended to weaker infinite Dedekind-finite sets? Suppose $L[A]=L$, as Joel points out in his answer this implies $A\cap L\in L$.

  1. Suppose $L[A]=L$ for every $A\in V$, is there something to say about $V$ and $L$? (in the sense that $V$ is somewhat minimal over $L$ (that is if it has non-well orderable sets, then this is the only difference of $V$ from $L$))
  2. Suppose $V$ is somewhat larger than the above description (for example, $V$ is the Feferman-Levy model in which $\omega_1$ is singular and the reals have cardinality $\aleph_1$), is there anything to say about sets for which $L[A]=L$? Can we in some sense generate a model $L\subseteq M\subseteq V$ which behaves as described above (some minimality property)?

Regarding your last question, it is easy to see that if $A\cap L=\varnothing$, then $L[A]=L$, because at every stage, if we have agreement $L_\alpha[A]=L_\alpha$ so far, then having $A$ as a predicate doesn't help us to define any new sets beyond the empty predicate (since the answer for whether a set in $L_\alpha$ is in $A$ is always no), and so $L_{\alpha+1}[A]=L_{\alpha+1}$.

More generally, for your title question, we have the following:

Theorem. $L[A]=L$ if and only if $A\cap L\in L$.

Proof. If $L[A]=L$, then clearly $A\cap L\in L$. Conversely, if $A\cap L\in L$, then one can show inductively that $L_\alpha[A]=L_\alpha[A\cap L]$, which is contained in $L$. That is, having $A$ as a predicate over $L_\alpha[A]$ is just as good as having $A\cap L$ as a predicate, if you know $L_\alpha[A]\subset L$. Since $A\cap L\in L$, this means that constructing relative to the predicate $A$ never leaves $L$. QED

Note however that the hypothesis $A\cap L\in L$ does not imply $L_\alpha[A]=L_\alpha$, because the predicate $A$ may allow you to define sets more quickly, even when they are in $L$.

This seems now to answer your conjecture:

Corollary. If $A$ is Dedekind finite (in particular, if $A$ is amorphous), then $L[A]=L$.

Proof. If $A$ is Dedekind finite, then $L\cap A$ must be finite, since otherwise we could construct a countably infinite subset using the $L$-order. Thus, $L\cap A\in L$, and so $L[A]=L$. QED

And your updated question 1 seems to be answered by:

Theorem. The following are equivalent:

  • $L[A]=L$ for every set $A$.
  • $V=L$.

Proof. If $V=L$, then clearly $L[A]=L$ for every $A$. Conversely, if $V\neq L$, then let $A$ be any $\in$-minimal set not in $L$. Thus, $A\cap L=A\notin L$, and so $L[A]\neq L$. QED

  • $\begingroup$ Of course! This is a stupid question. I am aware that if $M$ is a transitive model of $ZF$ and $A\cap M\in M$ then $L[A]\subseteq M$, in particular $A\cap L\in L$ implies that. I will restate my question in a clearer form, as I am more interested in the case of $A$ being iDf (or amorphous). Many thanks. $\endgroup$ – Asaf Karagila Jul 19 '11 at 23:40
  • $\begingroup$ I added a corollary, which seems to answer your conjecture. $\endgroup$ – Joel David Hamkins Jul 20 '11 at 0:02
  • $\begingroup$ I noticed that the corollary goes through for merely Dedekind finite, since if $A\cap L$ is infinite, then we can find a countably infinite subset using the definable L order. And the theorem at the end answers your new Question 1 at the end. $\endgroup$ – Joel David Hamkins Jul 20 '11 at 0:36
  • $\begingroup$ Joel: Thanks! After considering this for most of the day I figured out how the second question is answered by this theorem as well, negatively too. Since $L$ is absolute, if $M$ the class of all sets hereditary have the property $L[A]=L$ contains $L$ and is transitive, so by the theorem in your post we have that $L=M$. I'd think that if we do not require hereditary then this is no longer a transitive class, but I would expect a relatively simple argument to show that it is also not a model of ZF. $\endgroup$ – Asaf Karagila Jul 20 '11 at 21:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.