1
$\begingroup$

I've been playing with abstract functions and infinite sets and have come up with the concept of a "requisite function". Since the concept is relatively simple, if the concept is not flawed, then someone must have come up with this concept before and named it. The question is, what is the official name for this concept?

$A$ and $B$ are infinite sets. $f: A \to| B$ is a partial function such that some elements of $A$ are mapped to some elements of $B$.
Set $A'$ is the restricted subset of $A$ such that every element of $A'$ is mapped to some element of $B$.
A requisite function $f_r: A \to A'$ maps every element of $A$ to every element of $A'$. If an element of $A$ is already in $A'$, $f_r$ does not necessarily map the given element to itself.

Example 1:
Sets $A$ and $B$ are identical sets of all natural numbers. $f(x)=\frac x2=y$ such that $x \in A$ and $y \in B$. Therefore, $x$ must be divisible by $2$. Therefore, $A'$ is the set of all natural numbers that are divisible by $2$. Therefore, $f_r(x)=2x$.
This concept also extends to function powers: if $f^2(x)=\frac x {2^2}$ such that $x \in A$ and $y \in B$, then $x$ must be divisible by $2^2$. Therefore, the requisite function of $f^2(x)$ is $f^2_r(x)=2^2x$.
In general, if $f^n(x)=\frac x {2^n}$ then $f_r$ of $f^n(x)$ is $f^n_r(x)=2^nx$.

Example 2:
Sets $A$ and $B$ are identical sets of all natural numbers. $f(x)=x+1$ such that $x \in A$ and $y \in B$. Every element of $A$ is already mapped to some element of $B$, therefore $A=A'$, therefore $f_r(x)$ is just an identity function.

$\endgroup$
  • $\begingroup$ Do you mean the range of the function? (As contrasted with the codomain.) $\endgroup$ – Patrick Stevens Aug 21 '16 at 15:43
  • 1
    $\begingroup$ There are partial functions, which are relations $f: A \to B$ such that not all elements of $a$ are mapped, and there exists a subset $A' \subseteq A$ for which $f |_{A'}$ is a function. Don't know about these "requisite functions", though. $\endgroup$ – AJY Aug 21 '16 at 15:43
  • 1
    $\begingroup$ @Patrick Stevens Range of the function would be a subset of B, since B is the codomain. I mean a function that maps A to something like 'inverse range'. $\endgroup$ – ZyTelevan Aug 21 '16 at 15:49
  • $\begingroup$ Sounds like you are just looking for a surjective (a.k.a. onto) function from an infinite set $A$ to a subset $A'$. $\endgroup$ – Rahul Aug 21 '16 at 15:51
  • $\begingroup$ Can't you just pick an arbitrary element $a\in A'$ and define $f(x) = x$ if $x\in A'$ and $a$ otherwise? That satisfies your stated conditions. $\endgroup$ – Rahul Aug 21 '16 at 15:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.