MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Given language $L$. $P$ is a 1-place predicate in $L$. Let language $L_0 = L \setminus \{P\}$. Let $\sigma$ be a sentence of $L$ (may contain symbol $P$). $\mathfrak{A}$ is a structure of $L$, and $\sigma$ is true in $\mathfrak{A}$.

Assume for all structure $\mathfrak{B}$ of $L$, if $h: |\mathfrak{A}|\to|\mathfrak{B}|$ witnesses that $\mathfrak{B}|L_0$ (B restricted to L_0) and $\mathfrak{A}|L_0$ are isomorphic and $\sigma$ is true in $\mathfrak{B}$, then $h$ also witnesses that $\mathfrak{B}$ and $\mathfrak{A}$ (without the restriction) are isomorphic, i.e. $h[P^\mathfrak{A}]=P^\mathfrak{B}$.

Then can we conclude that $P^\mathfrak{A}$ is a definable subset in structure $\mathfrak{A}|L_0$?

The intuition is that if $P^\mathfrak{A}$ is somewhat determined by $\sigma$, can we then turn $\sigma$ into a definition sentence?

When I was tring to prove that $P^\mathfrak{A}$ may not be definable, I found both the automorphism and the cardinality arguments don't work. Have I missed anything?

share|cite|improve this question
Probably you intent to assert $\sigma$ in $\mathfrak{A}$ as well as $\mathfrak{B}$? Otherwise the only way your hypothesis can be true is if the structures are empty. – Joel David Hamkins Mar 26 '13 at 17:00
Joel, the end of the first paragraph says $\sigma$ is true in $\mathfrak A$ (and MO is not showing me any edit history to indicate this was added after your comment). I'm voting to close because, as Noah pointed out, the answer is a well-known theorem. – Andreas Blass Mar 26 '13 at 19:18
Ah, I had missed that. – Joel David Hamkins Mar 26 '13 at 20:47

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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