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Every parameter-free definable set is preserved setwise by every automorphism of a first-order structure. For atomic formulas, preservation follows from the definition of a structure automorphism. Logical connectives preserve that equivalence, and a bijection transports existential and universal witnesses in both directions. Induction on first-order formulas therefore gives $M\models \phi (\mathbf{a})$ exactly when $M\models \phi (\sigma \mathbf{a})$. Applied to a defining formula, this proves the displayed equality. With parameters from $B$, restrict to automorphisms fixing $B$ pointwise. An automorphism moving a proposed singleton or taking a proposed function-graph point outside the graph is consequently a direct obstruction to definability.

A function $f:M^n\to M$ is a definable function if its graph of a function is a definable set. Equivalently, a first-order formula $\theta (\mathbf{x},y)$ defines exactly the pairs with $y=f(\mathbf{x})$. Totality and uniqueness mean $M\models \forall \mathbf{x}\exists !y\theta (\mathbf{x},y)$. The same definition with restricted domains gives definable partial functions. A function need not be a symbol of the language to be definable; the graph can be specified through the existing relations and operations.

A definable set in a first-order structure $M$ is the set of tuples satisfying one first-order formula, possibly with a specified finite tuple of parameters $\mathbf{b}$ from the structure. A set definable over $B\subseteq M$ uses parameters from $B$. Parameter-free definability uses no extra named elements. The distinction matters: any singleton becomes definable if its own element is allowed as a parameter, while many singletons are not definable without parameters. Finite Boolean combinations of definable sets are definable, using logical conjunction, logical disjunction and logical negation.

Let $M$ be a countable omega-saturated model and $D\subseteq M$ finite. Then $D$ is a parameter-free definable set exactly when every structure automorphism preserves it setwise. One direction is automorphism invariance of parameter-free definability. For the converse, saturation implies homogeneity and the back-and-forth method show that equal empty-set types are in the same automorphism orbit. For $a\in D$, if no formula in $\mathrm{tp}(a)$ has its realization set contained in $D$, the partial type $\mathrm{tp}(a)\cup \{x\ne d:d\in D\}$ is finitely satisfiable. Its parameters form a finite set, so saturation realizes it outside $D$, contradicting invariance. Choose a defining neighbourhood for each $a\in D$ and take their finite disjunction. The empty set is defined by $x\ne x$. Setwise invariance permits automorphisms to permute points within $D$.

A definable set has parameter-free definability when its defining first-order formula uses only the original first-order language, with no additional constants naming elements. Constants already present in the language are allowed. Every such set is invariant under all automorphisms of a first-order structure. For example, zero in the constant-free additive integers is defined by $x+x=x$, while the singleton containing one is not parameter-free definable because negation is an additive automorphism moving it. Invariance is a necessary condition in every structure, and becomes sufficient for finite sets in a countable omega-saturated model.

Here parameter-free definability is intended: parameters are not silently added to the first-order language. A subset $D\subseteq A^n$ is a definable set when some $L$-first-order formula $\phi (x_1,\dots ,x_n)$ satisfiesA definable function is one whose graph is a definable set. Thus $f:A^n\to A$ is definable when an $L$-formula $\theta (\mathbf{x},y)$ satisfiesIn particular $\mathcal{A}\models \forall \mathbf{x}\exists !y\theta (\mathbf{x},y)$. Definability over a parameter set $B$ permits parameters from $B$ in these first-order formulas, equivalently new constants naming those parameters. That is a different requirement: allowing a parameter naming an element would immediately define its singleton.

Here fixed means fixed setwise: $\sigma [D]=D$ for every structure automorphism $\sigma$, not that each point must be fixed individually. A parameter-free definable set has this invariance by automorphism invariance of parameter-free definability.

For the converse, first establish the needed saturation implies homogeneity step. In a countable omega-saturated model, a finite partial elementary map can be extended to include any prescribed additional element in its domain. Transfer the element's complete type over the old finite domain along the map. Each finite part of this transported type has a witness because the old map is elementary; the omega-saturated model realizes the full transported type. The inequalities in the type maintain injectivity. Apply the same construction to the inverse map to enlarge the range. Alternating these steps through an enumeration of the countable domain is the back-and-forth method, giving a structure automorphism. In particular, any two elements with the same complete type over the empty set are related by a structure automorphism.

Assume now that $D$ is finite and invariant, and fix $a\in D$. All realizations in $A$ of $p_a=\mathrm{tp}(a/\varnothing )$ belong to $D$, by the preceding homogeneity and the invariance assumption. Suppose that no parameter-free first-order formula ${\theta }_a\in p_a$ has all its realizations in $D$. Then every finite subset ofis satisfiable in $A$: conjoin the finitely many first-order formulas from $p_a$, and use the assumed absence of a defining neighbourhood inside $D$ to find a witness outside $D$. This is a partial type over the finite parameter set $D$. The omega-saturated model would realize it by an element outside $D$ with the same empty-set type as $a$, a contradiction. Hence such a ${\theta }_a$ does exist.

There are only finitely many $a\in D$, soFor $D=\varnothing$, use $x\ne x$. Thus finite invariant sets in a countable saturated structure are definable. Finiteness is used twice: the extra parameters form a finite set, and the final disjunction is a first-order formula rather than an infinite disjunction.