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While it is easy to understand what a function is intuitively, I've been trying to wrap my head around how to precisely define what a function is using only mathematical notation. My attempt at this is below, but here is my preliminary understanding:

  1. A function can have multiple inputs or parameters, but it generates a single output
  2. Each output is unique for the input values provided.

Here is my attempt at a definition:

A relation $R \subset (D \times C)$ is a function if: $$ (\forall (d_1, c_1) \in R)(\forall (d_2, c_2) \in R)(d_1 = d_2 \rightarrow c_1 = c_2)$$

This definition should cover all functions, not just functions with one input, as $d_1$ and $d_2$ could be n-tuples that define the n inputs of a function, as every element of $R$ is actually an ordered pair $((x_1, x_2, ..., x_n),c)$

Does this look like a correct and precise definition? Or could it be written better? I couldn't find any formal definition of a function on the web, even on Wikipedia.

Finally, is it correct to say that all functions with n inputs are (n+1)-ary relations? Since $((x_1, x_2, ..., x_n),c)$ is the same as $(x_1, x_2, ..., x_n,c)$.

Thanks.

Jean-Claude Arbaut
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esotechnica
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  • Try to Google what it Means for a map to be "well defined". I think this will be something along the line you are looking for. – An.Ditlev Sep 07 '16 at 07:08
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    Wikipedia does give a definition that should be sufficient. https://en.wikipedia.org/wiki/Function_(mathematics)#Definition – Thanassis Sep 07 '16 at 07:10
  • @Thanassis while it may be descriptively sufficient, I'm only interested in a symbolic definition using set theory and logic – esotechnica Sep 07 '16 at 10:22

2 Answers2

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The usual way to define a function, you are correct, is by relations. However, your definition is not ok, because for example, an empty relation fits your definition.

The correct definition must say something like:

A relation $R\subset A\times B$ is a function if each element $a$ of $A$ is contained in exactly one relation $(a,b)\in R$

In pure terms, you can write this as

$$\forall a\in A\exists! b\in B: (a,b)\in R$$

Where the $\exists!$ quantifier means "exists precisely one". A longer version (with only $\exists$ and $\forall$) would be

$$\forall a\in A \exists b\in B:((a,b)\in R\land \forall b'\in B:(a,b')\in R\implies b=b').$$

Answer to your other question about functions on multiple inputs:

No, functions are always binary relations. A function of "multiple (say, $n$ inputs)" is actuall a function whose domain is a cartesian product. So, for example, a function of $n$ real numbers is in fact a function whose domain is equal to $\mathbb R^3$.

This means that technically, we shouldn't write $f(x,y)=xy$, for example. We should write $f((x,y))=xy$, because $f$ takes only one input, and that input is a tuple of two numbers.

5xum
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    Now that you mention it, is there any reason why you cant have a function $f:\emptyset \to A$? Other than that it would be useless? – pancini Sep 07 '16 at 07:12
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    @ElliotG No, there is no reason. In fact, $\emptyset$ is a function between $\emptyset$ and $A$ (since $\emptyset\subseteq \emptyset\times A$ is clearly true, and $\forall a\in \emptyset:(...)$ is also true no matter what the brackets hold. But if $X$ is non-empty, then $\emptyset$ cannot be a function between $X$ and any other set. – 5xum Sep 07 '16 at 07:16
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    Actually, to define a function you need at least the pair $(B,R)$ (and usually, it's defined using the tuple $(A,B,R)$, for simplicity I guess), so that the definition of a surjective function can make sense. Otherwise, you always let $B=Im(f)$ and all functions are surjective. – Jean-Claude Arbaut Sep 07 '16 at 07:44
  • @Jean-ClaudeArbaut Well, $R$ is a subset of $A\times B$, so you don't need to then once more define what $B$ is. Using the definition I described above, a surjective function can be simply defined. – 5xum Sep 07 '16 at 07:45
  • @5xum Given a subset of $A\times B$, you don't know for sure what $B$ is. For instance, $\Bbb N\times \Bbb N$ is a subset of $\Bbb N\times \Bbb Z$, but also of $\Bbb N\times \Bbb Q$, $\Bbb N\times \Bbb R$, and of course of itself, and many, many other sets. However, given the condition to be a function, you know what $A$ is. – Jean-Claude Arbaut Sep 07 '16 at 07:47
  • @Jean-ClaudeArbaut Ah, yes, you are right. – 5xum Sep 07 '16 at 07:48
  • @5xum thanks for the clear explanation, but I am just wondering, isn't an ordered pair that consists of an n-tuple and a single value equivalent to an (n+1)-tuple? (https://en.wikipedia.org/wiki/Tuple#Tuples_as_nested_ordered_pairs). Therefore a binary relation such as $A^n \times B$ would also be an (n+1)-ary relation? – esotechnica Sep 07 '16 at 10:41
  • @esotechnica Yes, if you define a tuple as a nested ordered pair, then the two things are the same. But you would still say that the function maps from $A^n$ (one set) to $B$ (one set), so it is still, in some way, binary. But don't get too hung up on the details. You understand the concept, and that's more important than knowing exactly which box you have to put it in. – 5xum Sep 07 '16 at 10:43
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5xum is right -- your definition is not sufficient. However, you may be interested to know that your definition,

A relation $R \subset (D \times C)$ is a function if: $$ (\forall (d_1, c_1) \in R)(\forall (d_2, c_2) \in R)(d_1 = d_2 \rightarrow c_1 = c_2)$$

is something we care about in some branches of math! This is called a partial function from $D$ to $C$ (sometimes denoted $f: D \rightharpoonup C$).

Partial functions are especially used in the theory of recursive functions (or computable functions from $\mathbb{N}$ to $\mathbb{N}$).

Caleb Stanford
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