@DawoodibnKareem It is a huge assumption and it is actually bogus, as you might have suspected. Per that bogus assumption, let W be the set of all well-defined sets. For each member S of W, it is well-defined and so "S∈S" has a well-defined truth-value. So let R = { S : S∈W ∧ S∉S }, which is a well-defined set because for every object S we have that "S∈W" has a well-defined truth-value, and so "S∈W ∧ S∉S" too has a well-defined truth-value (since either S∉W, or S∈W and "S∉S" is well-defined).

Thus "R∈R" has a well-defined truth-value. What is it?

@cmaster-reinstatemonica: See above. Also note that Kevin is completely wrong; just because one crazy way of constructing a model of PA uses ∅ does not at all imply that we cannot count if we do not have such as thing as ∅. I'm not saying ∅ is meaningless, but just that Kevin's reasoning is completely bogus.

@user21820 Good catch, I didn't see that. So, obviously, we do not have a well-defined one-element for the intersection operation. On the other hand, we absolutely must include the empty set to close the intersection operation: Consider a set A = {1} and a set B = {2}. The intersection A∩B = {} must yield the empty set. We need to include this in the theory, lest we be left with a hole in the definition of A∩B.

Any yes, I agree that there are other possible definitions for the natural numbers, which do not even touch on set theory.

@DoubleKnot Indeed, I was thinking of something like a top and bottom element. However, user21820 correctly pointed out that the top element cannot exist (the set of well-defined sets is not well-defined, contrary to my gut feeling :-( ). Nevertheless, we need the empty set to give meaning to {1}∩{2}. So, while we cannot have (and do not need) the set of well-defined sets, we do need the empty set, and not just because it's the zero element for the union operation.

@cmaster-reinstatemonica see, you've clarified yourself again... Set theory is full of logic traps especially for the "empty" and "all" edge cases if one only comprehends a set as a container casually. Now you invokes "completeness" no hole demand for all versions of it which itself is a "practical" stance. Even in the usual real number Ring R, division by 0 has a hole... Anyway glad you learned something new today...