Predicates and mathematical objects

Question

I'm reflecting about mathematical objects as numbers, sets/classes, graphs and so on.

Any class correspond to a predicate in one variable and any graph correspond to a predicate in two variables. There are exceptions for the correspondence in the other direction since "$x\notin x$" doesn't correspond to a class and "$R(x,y)\leftrightarrow \neg R(y,y)$" normally doesn't correspond to a graph.

In my opinion it seems to exist possible mathematical objects for each $n$ and predicates $p(x_1,...,x_n)$ with similar semi-correspondence. My intuition about this is that these objects are built upon the idea of simplex and I would like to know if it is true that

It always exists predicates $p_1,\dots,p_{n}$ such that
$p(x_1,\dots,x_n)\leftrightarrow\wedge_{i=1}^{n}p_i(x_1,\dots,x_{i-1},x_{i+1},\dots,x_n)$

Answer 1

Before addressing your actual question, let me point out that there are serious issues with your motivating examples.

First of all, "$x\not\in x$" does correspond to a class - namely, the universal class (in ZFC at least; a smaller class under NF). Every formula defines a class; that's what classes are, really.

Meanwhile, the issue with the $R$ you describe is that you haven't really (but see the bulletpoint below) defined a relation, but rather written a property a relation may or may not have. That's the issue with a self-referential "definition" (note that $R$ occurs on both sides of the expression you've written: when defining an object you need to show that it $(1)$ holds of something and $(2)$ only holds of one thing, and you haven't done that. Actually, this is something you have to do with any purported definition, but self-reference in particular makes $(1)$ especially nontrivial. (See also my old answer here.)

• Interestingly, your property does pin down a unique relation on a unique domain (= set of "vertices"): setting $x=y$ shows that we can't have any vertices at all, and there's only one binary relation on the emptyset. However, that probably wasn't intended, and indeed you would generally also have to define the domain of the graph. So when I said above that you haven't really defined a relation, what I really meant was that you haven't shown that you have defined a relation.

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These issues, however, don't actually impact your question (although they probably do impact its motivation).

Your question has an easy negative answer: let $p(x_1,...,x_n)$ hold iff for some $i\in\{1,...,n\}$ we have $x_i=0$. Then for any tuple of objects $a_1,...,a_n$ we have $$p(0,a_2,a_3,...,a_n)\wedge p(a_1,0,a_3,...,a_n)\wedge...\wedge p(a_1,a_2,..., 0).$$ So the desired $p_i$s would have to satisfy $$p_1(a_2,a_3,...,a_n)\wedge p_2(a_1,a_3,...,a_n)\wedge ...\wedge p_n(a_1,a_2,..., a_{n-1})$$ for every tuple of objects $a_1,a_2,...,a_n$. But then we have $$\bigwedge_{i\in\{1,...,n\}}p_i(a_1,..., a_{i-1}, a_{i+1}, ..., a_n)$$ for every tuple of objects $a_1,..., a_n$, which is a problem since e.g. $\neg p(1,1,...,1)$.

Incidentally, it's easy to see that the only possible candidate $p_i$s, for a given $p$, are $$p_i(a_1,..., a_{n-1})\iff \exists b(p_i(a_1, a_2,..., a_{i-1}, b, a_i, ..., a_n)).$$