Preliminaries

Let ${\displaystyle f:X\to Y}$ be a map. Given any subset ${\displaystyle S\subseteq X,}$ define its image under ${\displaystyle f}$ to be the set:

$${\displaystyle f(S):=\{f(s)~:~s\in S\}}$$

and define its preimage or inverse image under ${\displaystyle f}$ to be the set:

$${\displaystyle f^{-1}(S):=\{x\in X~:~f(x)\in S\}.}$$

Given ${\displaystyle y\in Y,}$ the fiber of ${\displaystyle f}$ over ${\displaystyle y}$ is defined to be the preimage:

$${\displaystyle f^{-1}(y):=f^{-1}(\{y\})=\{x\in X~:~f(x)=y\}.}$$

Any preimage of a single point in ${\displaystyle f}$'s codomain ${\displaystyle Y}$ is referred to as a fiber of ${\displaystyle f.}$

Saturated sets

A set ${\displaystyle C}$ is called ${\displaystyle f}$-saturated and is said to be saturated with respect to ${\displaystyle f}$ if ${\displaystyle C}$ is a subset of ${\displaystyle f}$'s domain ${\displaystyle X}$ and if any of the following equivalent conditions are satisfied:^[1]

1. ${\displaystyle C=f^{-1}(f(C)).}$
2. There exists a set ${\displaystyle S}$ such that ${\displaystyle C=f^{-1}(S).}$
   • Any such set ${\displaystyle S}$ necessarily contains ${\displaystyle f(C)}$ as a subset and moreover, it will also necessarily satisfy the equality ${\displaystyle f(C)=S\cap \operatorname {Im} f,}$ where ${\displaystyle \operatorname {Im} f:=f(X)}$ denotes the image of ${\displaystyle f.}$
3. If ${\displaystyle c\in C}$ and ${\displaystyle x\in X}$ satisfy ${\displaystyle f(x)=f(c),}$ then ${\displaystyle x\in C.}$
4. If ${\displaystyle y\in Y}$ is such that the fiber ${\displaystyle f^{-1}(y)}$ intersects ${\displaystyle C}$ (that is, if ${\displaystyle f^{-1}(y)\cap C\neq \varnothing }$), then this entire fiber is necessarily a subset of ${\displaystyle C}$ (that is, ${\displaystyle f^{-1}(y)\subseteq C}$).
5. For every ${\displaystyle y\in Y,}$ the intersection ${\displaystyle C\cap f^{-1}(y)}$ is equal to the empty set ${\displaystyle \varnothing }$ or to ${\displaystyle f^{-1}(y).}$

Related to computability theory, this notion can be extended to programs. Here, considering a subset ${\displaystyle A\subseteq \mathbb {N} }$, this can be considered saturated (or extensional) if ${\displaystyle \forall m,n\in \mathbb {N} ,m\in A,\vee \phi _{m}=\phi _{n}\Rightarrow n\in A}$. In words, given two programs, if the first program is in the set of programs satisfying the property and two programs are computing the same thing, then also the second program satisfies the property. This means that if one program with a certain property is in the set, all programs computing the same function must also be in the set).

In this context, this notion can extend Rice's theorem, stating that:

Let ${\displaystyle A}$ be a subset such that ${\displaystyle A\neq \emptyset ,A\neq \mathbb {N} ,A\subseteq N}$. If ${\displaystyle A}$ is saturated, then ${\displaystyle A}$ is not recursive.