Do functions require set theory?

Question

Do functions require set theory in order for us to even use them or talk about them?

Answer 1

Rigorously, it can. A function can be defined as an ordered triple: $(D,R,S)$, where $D$ is a set called the domain, $R$ is a set called the range, and $S$ is a set of ordered pairs that maps inputs to outputs.

Here's an example of a function:

$$( \{a,b,c\}, \{1,2,3,4,5,6,7,8\}, \{(a,3),(b,5),(c,3)\} )\text{.}$$

An ordered pair is, itself, defined as a set. The ordered pair $(a,b)$ is defined as $\{ \{a\}, \{a,b\} \}$.

So, rigorously, (contemporarily) yes; a function requires set theory. (See Wikipedia for the definition of a function. See Calculus by Spivak for the definition of an ordered set).

However, people don't tend to think at this level when talking about functions. We explain functions to children in middle school. But we don't go through the process of defining them rigorously at the level I have in this answer. So it seems that we can talk about them and use them without knowledge of set theory.

Answer 2

This depends on what foundational system you're using.

If you're using a set theory then everything is a set, including functions, and the only way to talk about a mathematical object is to construct it as a set. In this approach a function $f$ is usually implemented as a set of ordered pairs, where $(x,y)\in f$ is then interpreted as $f(x)=y$. This is by far the most common approach and the only one you're likely to encounter unless you go into foundational math.

This is not the only option however, there are different foundational systems in which functions are primitive notions! An example of such a system is (homotopy) type theory, in which given two types $A$ and $B$ one can form the function type $A\to B$ whose elements are functions and are not defined in terms of relations or other previously established notions.