Let X be a nonempty set. A sequence (of X) is a function s: N -> X. It is customary to write xn for s(n), so that we can think of a sequence as X0, X1, X2, ...
In this exercise, we will show that a sequence X0, X1, X2, ... satisfying X2 ⊆X1 ⊆X0 cannot exist, and therefore conclude that no set can contain itself.
(a) Let f: X -> Y be a function. Use the Axiom of Comprehension to show that R(f) = {y ∈ Y: ∃x ∈ X, f(x) = y} is a set. We call the above the range of f. Use this to conclude that for a given set X and sequence s, R(s) = {x ∈ X: ∃n ∈ N, (n, x) ∈ s} is a set.
(b) Use the Axiom of Foundation to argue that R(s) has a smallest element in terms of set containment (that is, there is x ∈ R(s) such that for all y ∈ R(s), y ⊆x).
(c) Conclude that there cannot exist a sequence of the form X2 ⊆X1 ⊆X0.
(d) Use this to conclude that there cannot be a self-containing set (i.e., x such that x ∈ x). (Hint: Suppose such an x exists. Consider a sequence s on {x} to arrive at a contradiction).
