Do functions require set theory?

Question

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

It depends on what level you are discussing functions. Technically, a function is defined as a thing that sends something from one set to another. It really depends on the context. In algebra/precalculus, the only functions studied are functions from reals to reals like the trig functions, polynomials, and exponentials. But later on in mathematics, when you get into analysis, other types of functions are discussed, and they are analyzed based on their set-theoretic properties. — clathratus, Jun 23 '19 at 03:59
To the down-voters: perhaps you should comment your reasons for down-voting; letting people know what is wrong with their post is better for the site. — clathratus, Jun 23 '19 at 04:14
It depends. The core notion of function is that if you give it an object of the right type (input type) then it will produce an output of some type (output type). But the core question is what you are allowed to construct to be functions. If you only permit lambda abstraction, then you do not need to base your foundations on set theory, but you will lack certain abilities. For example, even if you permit partial functions and can use induction on any property including involving well-definedness (termination), this does not enable you to construct some functions that you can construct in ZFC. — user21820, Jun 30 '19 at 17:38

score 4 · Answer 1 · answered Jun 23 '19 at 09:07

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.

Is an ordered pair in itself a set? If so, I thought sets didn't have order? — user557009, Jun 23 '19 at 13:46
Indeed, sets are not ordered, but as I said everything needs to be implemented as a set. The usual way to code the ordered pair $(a,b)$ as a set is the so called "Kuratowski pair", where $(a,b)$ is defined to be the set ${{a},{a,b}}$. (there are other possibilities and sometimes different ways to code a pair are actually useful). Incidentally note that the ordered pair is also a primitive notion in type theory! — Alessandro Codenotti, Jun 23 '19 at 13:51

score 1 · Answer 2 · edited Jun 23 '19 at 05:32

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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.

edited Jun 23 '19 at 05:32

Henno Brandsma

242,131

answered Jun 23 '19 at 04:02

NicNic8

6,951

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Mathematics does not need to be based on set theory. Lawvere created a framework were the basis of mathematics is not set theory, but category theory; there, “arrows” are a primitive concept. “Rigorously, yes” presupposes a particular foundation for mathematics. – Arturo Magidin Jun 23 '19 at 04:05
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@ArturoMagidin Pardon me if I seem ignorant (I know nothing about category theory), but isn't set theory the foundation for basically all contemporary mathematics? – clathratus Jun 23 '19 at 04:09
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@ArturoMagidin your comment seems to conflate the concept of arrow and function. What then is an arrow in the category of sets? – Infimum Jun 23 '19 at 04:38
Actually an axiom of set theory itself, the axiom of replacement, makes use of a notion of function that is not defined through sets, but comes directly from first-order logic. – celtschk Jun 23 '19 at 05:16
@clathratus A satisfactory foundation in the eyes of most, probably. But calling it the foundation is a bridge too far. Both because it presupposes that such a thing would be unique, and also because there is some diversity of opinion about whether the standard set theoretical foundations are the best approach, or if other options should be explored. – spaceisdarkgreen Jun 23 '19 at 05:21
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@clathratus: While mathematics is commonly presented as based on set theory these days, it is not the only possible way to base mathematics. – Arturo Magidin Jun 23 '19 at 05:31
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@Infimum: I’m not confusing things; the question was whether you needed to have set theory to be able to talk about functions. No, you do not, among other things because you can base mathematics on things other than set theory. – Arturo Magidin Jun 23 '19 at 05:32
@ArturoMagidin Is it just those two? Functions in terms of set theory or functions in terms of category theory? – user557009 Jun 23 '19 at 05:45
@user557009: No, those aren't the only ones. There are many different ways to found mathematics; Set Theory and Category Theory just happen to be the two most developes. And as celtschk notes, in some formulations of set theory, the idea of a function actually precedes set theory, as it is required to formulate axioms like Replacement. – Arturo Magidin Jun 23 '19 at 05:49
I am lost then, which one is which? Is there a name for the formulation that has functions defined through sets? Or one where functions are defined through logic? Or categories? How many "pathways" are there? – user557009 Jun 23 '19 at 05:54

Do functions require set theory?

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