Question

Suppose one has a bunch of possible elements, x, that satisfy some parameter P. Then suppose one tries to form a set, X, that includes all satisfiers of P, i.e., all those x. Assume that X doesn't itself satisfy P. So, if X is considered an element of itself, it is not because of satisfying P. It is manually inserted into itself, so to speak, when such insertion occurs. Consequently, X contains itself and all the x; let the x be noncircular sets. Now, X is the set of all noncircular sets, is itself circular, but doesn't generate the Russell paradox.

Even without such a manual insertion procedure, could we jerry-rig the intensional parameters for elementhood here, by changing P from noncircular to noncircular or identical to X? If an element x is noncircular or when x = X, then x is an element of X. Again, we have a set containing all noncircular sets but which also contains itself without generating the Russell paradox (X is not both inside and outside itself but is only inside itself. Yet, its interiority is not by satisfying the parameter of exteriority, so it is not both interior and exterior to itself). Is there some sort of revenge version of that paradox that can be formulated here? Are disjunctive properties omitted from the domain of even the naive comprehension principle? Suppose one has a set of possible elements, x, satisfying some parameter P. What happens when trying to form a set X that includes all satisfiers of P?

A. X is defined as a circular set that generates the Russell paradox.

B. X contains itself, and all noncircular sets, but avoids the Russell paradox.

C. X satisfies parameter P, leading to a contradiction.

D. The set X is guaranteed to be noncircular and free from paradoxes.

Answer 1

X contains itself and all the x, and changing the parameter P does not generate a revenge version of the Russell paradox.

Step-by-step explanation:

X contains itself and all the x

In this scenario, X is defined as a set that includes all elements x that satisfy some parameter P. However, X itself is not included as an element of X because it is manually inserted into itself, separate from satisfying P. So X contains itself and all the x, with the x being noncircular sets.

Changing the parameter P

Even without the manual insertion procedure, if we change the parameter P from noncircular to noncircular or identical to X, then an element x that is noncircular or x = X would be considered an element of X. This would result in X being a set that includes all noncircular sets and also contains itself, but without generating the Russell paradox.

Revenge version of Russell paradox

In this case, there isn't a revenge version of the Russell paradox that can be formulated. The disjunctive properties are not omitted from the naive comprehension principle, and the set X is guaranteed to be noncircular and free from paradoxes.