(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?


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.

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

  • 1
    $\begingroup$ 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! $\endgroup$ – Alex Kruckman Mar 3 '17 at 3:02

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