# Definability vs Automorphisms

(I am skipping any setup stuff and speaking roughly)

One fact that I am sure of is that a definable subset $X$ is fixed by all automorphisms of the (super)structure.

I simply wonder the converse:

"If a subset $X$ is fixed by all automorphisms, then $X$ is a definable set."

Is this true (in general)? If not, could you provide some counterexample and the cases/conditions that the latter assertion is true?

## 2 Answers

Look at the naturals under the usual addition, multiplication. The only automorphism is the trivial one, but there are many undefinable subsets.

The same is true of the reals.

• I think OP may have meant to ask about automorphisms of arbitrary superstructures (including saturated ones). – tomasz Jun 10 '13 at 17:54
• The question may change, or there may be a new one. Right now I do not know how much background OP has. – André Nicolas Jun 10 '13 at 17:57

Of course this is true if your structure is big enough (sufficiently saturated), which is kind of assumed by putting this question into model theory. In-fact the following statement holds: A set $D$ is definable over $A$ iff it is fixed by all automorphism fixing $A$. to proof right to left note that the $D$-membership depends only on the type over $A$. Then check that $\{tp(d/D):d\in D\}=X$ is a clopen set in the type space to finish the proof (Check that $X$ and $S(A)-X$ are closed).

• This is an old answer to an old question (coming on 4 years old), but I have to point out that it's not correct. The correct statement is that if $M$ is saturated, and a definable set $X\subseteq M$ (meaning definable with any parameters from $M$) is fixed by all automorphisms fixing $A$, then $X$ is definable with parameters from $A$. The assumption that $X$ is definable is crucial for showing that $X$ is a clopen subset of $S(A)$. After all, for any subset $Y\subseteq S(A)$, the set $\{a\in M\mid \text{tp}(a/A)\in Y\}$ is $A$-invariant, but not every subset of $S(A)$ is clopen! – Alex Kruckman Mar 3 '17 at 3:02