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A K-ary relation is a class of K-tuples, for some positive integer K. For example, a binary relation is a class of ordered pairs.

A K-ary relation is also a subset of the cartesian product of the sets $X_{1}$, $X_{2}$, $X_{3}$, ... , $X_{K-1}$ and $X_{K}$.

This must mean that a class of K-tuples is equivalent to a subset of the cartesian product of $X_{1}$, $X_{2}$, $X_{3}$, ... , $X_{K-1}$ and $X_{K}$. For the purpose of presenting both definitions to my students, I would like to prove this, and I cannot do so.

Proving that if you have a subset of the cartesian product of $X_{1}$, $X_{2}$, $X_{3}$, ... , $X_{K-1}$ and $X_{K}$, then you must have a class of K-tuples seems fine, but I cannot prove the converse. That is, if you have a class of K-tuples, then you must have a subset of the cartesian product of $X_{1}$, $X_{2}$, $X_{3}$, ... , $X_{K-1}$ and $X_{K}$.

I know that a set of K-tuples (that all possess the same property) is equivalent to a subset of the cartesian product of $X_{1}$, $X_{2}$, $X_{3}$, ... , $X_{K-1}$ and $X_{K}$, so it seems like I just have to prove that a class of K-tuples is a set of those same K-tuples.

To prove this, am I going to have to show that a class of K-tuples satisfies the axioms of ZFC?

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    For the powers of $X$, use the math-notation $X^{k-1}$ to make them look like $X^{k-1}$. – kaba Mar 07 '25 at 00:07
  • I think your second paragraph should say that a k-ary relation is a subset of $X^k$ (i.e. does not include the lower powers). – kaba Mar 07 '25 at 00:10
  • I actually intended them to be subscripts: $X_{1}$, $X_{2}$, and so on. But thanks for the tip.

    I think you're correct that a k-ary relation is a subset of $X^{K}$ for homogeneous relations, but I'm talking about in the general case when $X_{1}$, $X_{2}$, and so on are different.

     –  Mar 07 '25 at 02:00
  • Does your question come down to: is a subclass of a set always a subset? – Alex Kruckman Mar 08 '25 at 00:24
  • If so: see here and here. – Alex Kruckman Mar 08 '25 at 00:26
  • "To prove this, am I going to have to show that a class of K-tuples satisfies the axioms of ZFC?" Certainly not! A set is not something that satisfies the axioms of ZFC. The axioms of ZFC tell us that certain sets must exist. For example, the empty set is obviously not a model of ZFC - instead, the axioms of ZFC prove that the empty set exists. – Alex Kruckman Mar 10 '25 at 13:57
  • Your problem is whether a certain "class" is a set, but then you talk about ZFC, in which everything is a set. In ZFC, "class" is just another name for "set", which is just another name for "object", where all three are just informal names for the values ZFC variables can take on. The definition of "$K$-tuple" you have uses "class" to be applicable to other set theories that allow proper classes. The definition of "$K$-nary relation" you have applies strictly to "set-only" set theories like ZFC. Remove that inconsistency, and there is no problem. – Paul Sinclair Mar 10 '25 at 17:32
  • @Paul Sinclair Thank you very much. So, if I am interpreting you correctly, the class of K-tuples definition is in the theory of, say, NBG, and the cartesian product definition is in the theory of, say, ZFC? –  Mar 14 '25 at 13:32
  • More likely, the definition of K-tuples was intended to apply to both, with the understanding that "class = set" in ZFC. But the cartesian product definition applies to ZFC, or to theories where you cannot form a cartesian product between proper classes. (With some theories - e.g., Russell-Whitehead - this is possible, but other theories have limited capacity for operations on proper classes. I don't recall enough about NBG to say for it.) – Paul Sinclair Mar 14 '25 at 15:09

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As written, it is false. { ⟨x,x⟩ : x=x } is a class of 2-tuples, but it is certainly not a set, not to say subset of anything. You need to check your definitions of "set" and "class" and "relation" very carefully. And you cannot say "S satisfies ZFC" if you actually mean "S is a set". This is why precision is fundamental if you want to truly understand logic.

user21820
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