# What is the math notation for this type of function?

A function that turns a real number into another real number can be represented like $f : \mathbb{R}\to \mathbb{R}$

What is the analogous way to represent a function that turns an unordered pair of elements of positive integers each in $\{1,...,n\}$ into a real number? I guess it would almost be something like $$f : \{1,...,n\} \times \{1,...,n\} \to \mathbb{R}$$ but is there a better notation that is more concise and that has the unorderedness?

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I have seen $[n]$ for $\{1,2,3,\ldots n\}$ but it is always defined, not considered standard like $\mathbb{R}$. You could extend your function to be on ordered pairs by symmetry, but maybe that obscures a point you want to make. –  Ross Millikan Oct 21 '11 at 16:04
you may also consider set of unordered pairs as a triangle $X = \{(i,j): 1 \leq i\leq j \leq n\}$ but it maybe also obscuring. –  Ilya Oct 21 '11 at 16:23
If $X$ is any set you may denote the set of unordered pairs of elements of $X$ by ${X\choose2}$. So your function can be described as $f:\ {[n]\choose 2}\to{\mathbb R}$. –  Christian Blatter Oct 21 '11 at 19:46
@Christian: I generally see ${X \choose 2}$ used to denote the set of subsets of $X$ of size $2$, which is very close but not quite the same thing. –  Qiaochu Yuan Oct 21 '11 at 20:23
@Qiaochu Yuan: I don't know whether you can call a singleton $\{a\}$ an "unordered pair". The OP will have to decide what he actually meant. –  Christian Blatter Oct 22 '11 at 20:01

The set $\{1, \ldots ,N \}$ is often written as $[N]$, so this could be $f: \operatorname{Sym}^2([N]) \to \mathbb{R}$. Here $\operatorname{Sym}$ means the symmetric product, that is, $\operatorname{Sym}^2(S)$ can be thought of as the set of unordered pairs of elements of $S$.

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interesting, I didn't know about either the [N] notation or the $Sym^2$ notation. –  opt Oct 21 '11 at 16:17
The $Sym$ notation is usually used with vector spaces, sometimes in conjuction with wedge notation (wedge is the anti-symmetric product). Using it with sets is a bit of abuse of notation. –  Craig Oct 21 '11 at 17:51

I would say that it might be best to preface your notation with a sentence explaining it, which will allow the notation itself to be more compact, and generally increase the understanding of the reader. For example, we could write:

Let $X=\{x\in\mathbb{N}\mid x\leq N\}$, and let $\sim$ be an equivalence relation on $X^2$ defined by $(a,b)\sim(c,d)$ iff either $a=c$ and $b=d$, or $a=d$ and $b=c$. Let $Y=X^2/\sim$, and let $f:Y\to\mathbb{R}$.

So, $Y$ can be thought of as the set of unordered pairs of positive integers up to $N$, and you can then proceed to use this notation every time you want to talk about such a function.

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I gave a +1 for your first sentence. –  Ｊ. Ｍ. Oct 22 '11 at 1:49

Looking at some of the comments I realize I might not know what you mean by ordered pair exactly. That said, under one way of interpreting the term "ordered pair" you simply can't usually pick a function at random from the set of all functions from elements of a sequence of the positive integers to the real numbers, such that it can get said to map from an unordered pair of elements of positive integers to a real number. For example, for the subtraction function - with this domain and codomain, it doesn't work out such that -(x, y)=-(y, x). The subtraction function takes ordered pairs such as (3, 4) to a real number. For a binary function B, you need commutation B(x, y)=B(y, x) to hold for the function to meaningfully take an unordered pair to a real number. For a trinary function T you need

T(x, y, z)=T(x, z, y)=T(y, x, z)=T(y, z, x)=T(z, x, y)=T(z, y, x). In general, for an n-ary function you'd need all permutations of variables to come as equivalent, which only holds in exceptional cases.

With N+ denoting the positive integers, you might write f:(N+, N+)->R, such that

for all x, y, f(x, y)=f(y, x). So, the function f takes a member m of N+, and another member n of N+ in that order, such that if m does not equal n, the order in which f took those elements of the domain doesn't matter since f(m, n)=f(n, m). The extra condition (axiom) of commutation or commutativity implies the unorderedness.

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@Doug: This is false; you very well can have a function taking unordered pairs of positive integers to real numbers, because the function need not have anything to do with the elements of the unordered pair. Given any non-empty set $X$, there are uncountably many functions from $X$ to $\mathbb{R}$. That doesn't change depending on whether $X$ is a set of unordered pairs of positive integers, or a set of anything else. So I think you are misinterpreting the question. –  Zev Chonoles Oct 21 '11 at 16:54
@Doug: A pair (ordered or unordered) of positive integers is one object. All functions input one object and output one object; whether or not those objects are pairs, triples, sequences, or whatever is immaterial. –  Zev Chonoles Oct 21 '11 at 23:45
You are confused, or rather, you are too rigid in your refusal to think about functions in any way other than the way you want to. I don't have time to continue discussing this. I recommend you ask a question about this issue, presumably the collective efforts of the community here can do a better job explaining this better than I can. –  Zev Chonoles Oct 22 '11 at 4:49