# What goes wrong if we explain Russell's paradox as resulting from an overly rigid link between property satisfaction and elementhood?

Specifically, consider the following: $$\exists c \exists m \forall x (x \notin c ⇒ (x \in m ⟺ P(x))) \land (x \in c \Rightarrow (x \notin m ⟺ P(x)))$$

We can call the above a "schema of premises" or "conjecture schema" (the phrase "conjecture schema" is to be thought of as analogous to the phrase "axiom schema").

Given that every value of x is either an element of c or not an element of c, the information provided by c and m should be enough to encode or represent everything about the intuitive conception of P(x) as a mapping from the value x to the truth value of P(x).

An answer -- that has unfortunately been deleted -- provided an example of a contradiction that can be deduced from the above conjecture schema (analogous to "axiom schema"), but the deduction relied upon some premises copied directly from ZFC, without adapting the ideas that motivated the formulation of the ZFC premises to the conjecture schema. Nevertheless, that answer at least provided some specific indications of reasoning, and that answer may have been of value in the process of developing a better answer.

What goes wrong with the conjecture schema? It is a simple enough proposal that something must at least appear to go wrong. Maybe an advantage of having at least two people looking at it is that it will be possible to confirm beyond any doubt that what appears to go wrong actually does go wrong.

• How about the other direction: something more complicated instead of something simpler? The proposed scheme at the beginning of this thread is a simplification that I initially thought was too simple and likely to lead to a contradiction. I started with the following ... Oct 24, 2019 at 5:24
• $\exists m \exists b \exists c \exists d \forall x [ [ x \in b \lor x \in c \lor x \in d] \land [x \notin b \lor x \notin c \lor x \notin d] \land [ [ x \in b \iff x \in c] \Rightarrow [x \in m \iff [P(x) \iff x \in d ] ] ] \land [ [ x \in b \iff x \in d] \Rightarrow [x \in m \iff [P(x) \iff x \in c ] ] ] \land [ [ x \in c \iff x \in d] \Rightarrow [x \in m \iff [P(x) \iff x \in b ] ] ]$ Oct 24, 2019 at 5:29
• "it's not clear what the motivation for the ZFC pairing axiom would be" Seriously? In what universe do you not want to be able to form pairs? Why does your pet principle take precedence over how sets are actually used? That's absolutely an example of something "going wrong" - what you have to throw out to accommodate it is far too much. Oct 26, 2019 at 23:59
• I think that your question is more appropriate for philosophy stack exchange and would get a better reception there. Oct 30, 2019 at 17:19

That axiom is satisfied by empty m,c and P(x) being a false statement.
Nothing goes wrong other than that axiom is vacuous and accomplishes nothing.

• If P(x) is $x \neq x$ then we want there to be a representation of P(x). In ZFC, we have the empty set corresponding to P(x). For the proposed scheme that is supposed to be more flexible, we also have a representation for P(x), although two sets are involved instead of just one. What is the difficulty that you see? Simply choosing P(x) that is false for all values of x doesn't seem to be a problem. What is the basis for your claim "vacuous" and your claim "accomplishes nothing"? Oct 23, 2019 at 22:59
• In Euclid's Elements, two constructed line segments are said to be in the same proportion to each other as two other line segments. That's a matter of definition, and it is said that the definition was proposed by Eudoxus. However, nowadays, the ratio of the lengths of two line segments is simply considered to be a single real number. Euclid uses two objects (analogous to the conjecture schema using c and m). So, from the point of view of having a single real number instead of the complicated approach to proportion in Euclid, can we assign the label "vacuous" to the approach in Euclid? Oct 25, 2019 at 11:33
• See the new answer here: math.stackexchange.com/a/4850001/710965 Jan 23 at 17:40

Introduce a defined binary relation $$[ \in ]$$ as follows:

$$h [\in] k \iff \exists t \exists u \left( k = (t,u) \land \left( \left( h \notin t \land h \in u \right) \lor \left( h \in t \land h \notin u \right) \right) \right)$$, where (t,u) is simply the ordered pair having t as first coordinate and u as second coordinate.

We have among our premises the following:

$$\exists c \exists m \forall x (x \notin c ⇒ (x \in m ⟺ \lnot (x [\in] x))) \land (x \in c \Rightarrow (x \notin m ⟺ \lnot (x [\in] x)))$$
In particular, plugging in $$x = (c,m)$$, we obtain:

$$\exists c \exists m ((c,m) \notin c ⇒ ((c,m) \in m ⟺ \lnot ((c,m) [\in] (c,m)))) \land ((c,m) \in c \Rightarrow ((c,m) \notin m ⟺ \lnot ((c,m) [\in] (c,m))))$$

Thus, by simple manipulation of connectives in propositional logic, we obtain:

To be seen again: $$\exists c \exists m \left[((c,m) \notin c \land (c,m) \in m ) \lor ((c,m) \in c \land (c,m) \notin m) \right] \iff \lnot ((c,m) [\in] (c,m))$$

However, if $$h = (c,m)$$ and $$k = (c,m)$$, then we can simply assign $$t=c \land u = m$$.

Thus, we have -- directly from our definition of $$[ \in ]$$ -- the following conclusion:

$$(c,m) [\in] (c,m) \iff \left[((c,m) \notin c \land (c,m) \in m ) \lor ((c,m) \in c \land ((c,m) \notin m) \right]$$

However, that conflicts with the statement that has the label "To be seen again."

Thus, we have a contradiction and the original conjecture schema or schema of premises is logically inconsistent, rather than vacuous.