I’ll try to present an argument in favor of the idea that a set that contains itself is a contradiction in terms.
However, in order to that, I’ll have to introduce a new term. I have to do this because I have trouble finding an existing term in mathematics that captures the concept that I want to express.
The notion that I am going to introduce will be represented by the expression “the volume of a set”.
obsrvr524 is using the word “size” to capture the same concept but that leads to a minor problem since the word “size” is already defined with respect to sets and it means something else – it refers to the number of elements. (A = {1, 2, 3, A}) has exactly (4) elements. No more and no less. The fact that (A) contains itself does not mean it has more than that.
So here we go. I’m going to kick it with a recursive definition.
Let the volume of a set be the sum of each member’s volume.
Let the volume of anything that is not a set be equal to (1).
For example, the volume of (B = {1, 2, 3}) is (3) because the set contains three elements, each one of which is not a set, which means, the volume of each element is exactly (1). Thus, the volume of (B) is (3 \times 1) which is (3).
The volume of (C = {1, 2, 3, {1, 2, 3}}), on the other hand, is equal to (6) even though its size (= cardinality) is equal to (4).
Let (n) be a number that represents the volume of (A = {1, 2, 3, A}).
What’s (n) equal to?
Let’s see. (A) contains (3) non-set elements and (1) set-element. This means the volume of (A) is (3 \times 1) plus the volume of the set-element it contains. We can simplify this to (3) plus the volume of the set-element. The set element it contains is (A) which means its volume is (n). Thus, the volume of (A) is (3 + n). This contradicts our earlier claim that the volume of (A) is (n). The volume of a set cannot be both (n) and (n + 3).