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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
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3 Answers

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
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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
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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
@Doug: Well, why not ask a question along the lines of "how can a function have more than one arity", explaining your point of view and asking whether what I am saying makes any sense. Don't you want to clear up why your post received all these downvotes, and why my comments have received all these upvotes? At any rate, I am sure the answers to the question will be enlightening (for both of us). Or, don't ask a question, and that will simply be the end of the matter - as I said, I am not going to discuss it further. –  Zev Chonoles Oct 22 '11 at 4:55
I see. On a few personal notes: I am aware you were suspended (I think) after/during a conversation with me. I did not flag any of your posts or desire your suspension at all, even if I believe you were consistently wrong. Second, when you post on a Q&A site, or communicate with others generally, there's implicitly some level of desire to understand others and be understood yourself. So downvotes tell you you're either wrong about something (which you should care about), or your audience doesn't understand what you've written or some issue at hand (which you should also care about) methinks. –  anon Oct 23 '11 at 0:30
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