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The empty set is an $n$-ary relation for every $n$, right?

How should we call a pair $(n;r)$ consisting of some number $n$ and an $n$-ary relation $r$?

To specify $n$ is necessary only when $r$ is empty, but because there are no reason for $r$ not to be empty, I need to specify $n$ explicitly.

Any term describing this situation?

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I've found the terms "figure" and "ground" in en.wikipedia.org/wiki/Theory_of_relations but Wikipedia here seems to use a non-standard terminology. –  porton Mar 26 '12 at 19:49
    
The empty set is a relation if and only if your definition of a relation is one that the empty set satisfies vacuously. –  Asaf Karagila Mar 26 '12 at 20:17

1 Answer 1

I don't believe that there is any common terminology for what you are asking about, but you should be careful.

Using the common recursive/inductive definition of $n$-tuples, where $$\langle a_0 , \ldots , a_{n-1} , a_n \rangle = \langle \langle a_0 , \ldots , a_{n-1} \rangle , a_n \rangle,$$ it follows that an $n$-tuple is also an $m$-tuple for all $m \leq n$. Therefore an $n$-ary relation is techincally also an $m$-ary relation for all $m \leq n$. (Differences may arise when talking about $n$-ary relations on a specific set $X$, but even here there might be ambiguity.)

Added due to comments below: Ignore the second paragraph.

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I define an $n$-tuple as a function of $n$ arguments, not inductively as you do, in order to match with my other definitions. There are no inherent trouble here, my question is purely terminological. –  porton Mar 26 '12 at 20:05
    
@porton: How do you define functions, then? –  Arthur Fischer Mar 26 '12 at 20:06
    
A function is a monovalued binary relation. If I describe it strictly, a binary relation is not a relation, because an ordered pair is (supposedly) not a tuple. Here there are some silliness dependent on a branch of math or CS, I know no way to define it in a non-silly way, every common way has it advantages and disadvantages. –  porton Mar 26 '12 at 20:10
    
@Arthur: This MO question might be interesting to your question. –  Asaf Karagila Mar 26 '12 at 20:16
    
@porton: I just wanted to know where your starting point was, to make certain that there was no circularity. –  Arthur Fischer Mar 26 '12 at 20:26

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