Let $L$ be the language of first-order logic with equality and $D$ a set, considered as a model of $L$. Is there a nice characterization of the subsets of $D^n$ that are definable in $L$ with parameters from $D$? For instance, presumably the definable subsets of $D^1$ are just the finite and cofinite subsets. (Any references appreciated too.)

  • $\begingroup$ I feel like you're asking for more than the definition of pointwise definability in a structure. There are sets which contain finite subsets that are not definable. Could you say a little bit more about what you mean? $\endgroup$ – Nika Dec 16 '19 at 23:49
  • $\begingroup$ I wanted simple necessary and sufficient conditions for a subset of $D^n$ to be characterized by a formula in $n$ variables and parameters in $D$. I was hoping they existed, since there is a simple characterization when $n=1$, namely the finite and cofinite subsets of $D$. $\endgroup$ – Andrew Bacon Dec 16 '19 at 23:55
  • $\begingroup$ @Nika, you mentioned that there are sets with finite subsets that are not definable. Keep in mind that this question is about which subsets of $D^n$ will be definable with parameters from $D$. $\endgroup$ – Andrew Ostergaard Dec 16 '19 at 23:56
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    $\begingroup$ "There are sets which contain finite subsets that are not definable". Every finite subset of $D$, $\{a_1...a_n\}$ is defined by the formula $x_1=a_1\vee ... \vee x_n=a_n$ with the parameters $a_1...a_n$. $\endgroup$ – Andrew Bacon Dec 16 '19 at 23:56
  • $\begingroup$ Whoops I must be misunderstanding something. I was thinking of an example in Enderton's A Mathematical Introduction to Logic which has $D = \{a,b,c\}$ and a relation on $D$, $E = \{\langle a,b \rangle, \langle a,c \rangle \}$, for which neither $\{ b \}$ nor $\{ c \}$ are definable. Did I get definable confused with definable with parameters in $D$? $\endgroup$ – Nika Dec 17 '19 at 0:02

Here is a proof that the theory of any set $D$ (over the empty signature) has quantifier elimination. By induction on formulas, it suffices to eliminate one existential quantifier at a time. That is, it suffices to prove that if $\varphi(x_1,\dots,x_n,y)$ is a quantifier-free formula then there exists a quantifier-free formula $\psi(x_1,\dots,x_n)$ such that $$D\models\forall x_1\dots\forall x_n(\exists y\varphi(x_1,\dots,x_n,y)\leftrightarrow \psi(x_1,\dots,x_n)).$$ To prove this, define the shape of $(a_1,\dots,a_n)\in D^n$ to be the equivalence relation $\{(i,j):a_i=a_j\}$ on the set $\{1,\dots,n\}$. Observe that if $(a_1,\dots,a_n)$ and $(b_1,\dots,b_n)$ have the same shape, there is an automorphism (i.e., bijection) $f:D\to D$ which satisfies $f(a_i)=b_i$ for all $i$. Thus, $D\models \exists y\varphi(a_1,\dots,a_n,y)\leftrightarrow \exists y\varphi(b_1,\dots,b_n,y)$. In other words, the truth of $\exists y\varphi(a_1,\dots,a_n,y)$ depends only on the shape of $(a_1,\dots,a_n)$.

Now for any equivalence relation $\sim$ on $\{1,\dots,n\}$, let $\psi_\sim(x_1,\dots,x_n)$ be a quantifier-free formula that expresses that $(x_1,\dots,x_n)$ is $\sim$-shaped (so $\psi$ is a big conjuction of formulas of the form $x_i=x_j$ or $\neg x_i=x_j$ depending on whether $i\sim j$). Let $\psi$ be the disjunction of $\psi_{\sim}$ over all $\sim$ such that $D\models\exists y\varphi(a_1,\dots,a_n,y)$ if $(a_1,\dots,a_n)$ is $\sim$-shaped. We then see that for each possible shape of $(a_1,\dots,a_n)\in D^n$, $D\models \exists y\varphi(a_1,\dots,a_n,y)\leftrightarrow \psi(a_1,\dots,a_n)$, and so $\psi$ has the desired property.

From quantifier elimination, it follows that every definable subset of $D^n$ is defined by a quantifier-free formula, which is just a Boolean combination of atomic formulas. There are just three types of atomic formulas (with parameters):

  • $x_i=x_j$
  • $x_i=d$ (or $d=x_i$) for some parameter $d\in D$
  • $d=e$ for some parameters $d,e\in D$.

In the first case the corresponding definable subset is $$\{(x_1,\dots,x_n)\in D^n:x_i=x_j\},$$ and in the second case the corresponding definable subset is $$\{(x_1,\dots,x_n)\in D^n:x_i=d\}.$$ In the third case it is either $D^n$ or $\emptyset$ depending on whether $d=e$ is true, so we can ignore that case. Thus the definable subsets of $D^n$ are Boolean combinations of the two types of sets above: "diagonal" subsets where two coordinates are equal, or "hyperplane" subsets where one coordinate has a fixed value.

When $n=1$ both of these types are finite or cofinite sets and so the definable sets are just the finite or cofinite sets. For $n>1$, there is not any description that is much simpler than "Boolean combinations of these sets". If you like, you could say that for any subset $A\subseteq D^n$ definable from parameters $d_1,\dots,d_m\in D$, there is a set $S$ of equivalence relations on $\{1,\dots,n+m\}$ such that $A$ is the set of all $(x_1,\dots,x_n)$ such that the shape of $(x_1,\dots,x_n,d_1,\dots,d_m)$ is in $S$.

  • $\begingroup$ Seeing the proof of quantifier elimination was really helpful for me. Thanks! $\endgroup$ – Andrew Bacon Dec 17 '19 at 1:09

If $D$ is finite, then we can define any subset of $D^n$ using parameters from $D$.

If $D$ is infinite, then it is what we call a strongly minimal structure: its definable subsets are indeed either finite or cofinite. The theory of $D$ is that of infinite sets, and this theory has quantifier elimination, which tells you something about the definable subsets of general $D^n$. The only atomic formula in this language is "$x = y$". This corresponds to a definable subset of the form $\{(x_1, \ldots, x_n) \in D^n : x_i = d\}$ for some $1 \leq i \leq n$ and $d \in D$. The definable subsets of $D^n$ are then Boolean combinations of such sets.

Edit: as Eric Wofsey pointed out in the comments, we can also use $\{(x_1, \ldots, x_n) \in D^n : x_i = x_j\}$ in our Boolean combination (getting for example the diagonal of $D^2$). The case of $\{(x_1, \ldots, x_n) \in D^n : d = e\}$, for $d, e \in D$ is really not interesting, because it is either the entire set or the empty set.

  • $\begingroup$ Thanks, this characterization is very helpful. Do you have a reference for the fact about quantifier elimination? $\endgroup$ – Andrew Bacon Dec 17 '19 at 0:07
  • $\begingroup$ And could you say a little more about why it implies that the definable sets have that form? Is there a general fact about theories with quantifier elimination you are appealing to? $\endgroup$ – Andrew Bacon Dec 17 '19 at 0:07
  • $\begingroup$ @AndrewBacon: This is immediate from the definition of quantifier elimination. Quantifier elimination says every formula is equivalent to a quantifier-free formula, and a quantifier-free formula is by definition a Boolean combination of atomic formulas. $\endgroup$ – Eric Wofsey Dec 17 '19 at 0:28
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    $\begingroup$ Oh, the description of the definable subsets corresponding to atomic formulas here is incorrect, though: they can also have the form $\{(x_1,\dots,x_n)\in D^n:x_i=x_j\}$. (Or $\{(x_1,\dots,x_n)\in D^n:d=e\}$ for some $d,e\in D$, but such a set is either $\emptyset$ or $D^n$ so we don't need to include that case.) This is because an atomic formula is $x=y$ where each of $x$ and $y$ can be either a variable or a parameter, so there are different cases depending on how many of them are parameters. $\endgroup$ – Eric Wofsey Dec 17 '19 at 0:34
  • $\begingroup$ Right, I can see how the definable subsets are Boolean combinations of those sets, given quantifier elimination. $\endgroup$ – Andrew Bacon Dec 17 '19 at 0:38

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