Hello folks,

I'm attempting some self-study in set theory using the text mentioned above. The exercises here are quite different from those in previous texts which I've used, so I was hoping I could present some of my attempts (so far, only from the first problem set) and receive some feedback as to whether it seems I am performing them correctly. First I'll briefly state the axioms they use:

Axiom of existence:

There exists a set which has no elements.

Axiom of extensionality:

If every element of X is an element of Y and every element of Y is an element of X, then X = Y.

Axiom Schema of Comprehension:

Let P(x) be a property of x. For any set A, there is a set B such that x ∈ B iff x ∈ A and P(x).

Axiom of pair:

For any A and B, there is a set C such that x is an element of C iff x = A or x = B.

Axiom of union:

For any set S, there exists a set U such that x ∈ U iff x ∈ A for some A ∈ S.

Axiom of Power set:

For any set S, there exists a set P such that X ∈ P iff X ⊆ S.

3.1: Show that the set of all x such that x ∈ A and x B exists.

3.1 solution: Let P(x, A, B) be the property x ∈ A and x B. Then W = {x ∈ A | P(x, A, B)} exists by the axiom schema of comprehension.

3.4 Let A and B be sets. Show there exists a unique set C such that x ∈ C iff either (x ∈ A and x B) or (x ∈ B and x A).

3.4 solution: By the axiom of pair, there is some set D such that x ∈ D iff x = A or x = B. Let P(x, A, B) be the property (x ∈ A and x B) or (x ∈ B and x A). Then W = {x

∈ D | P(x, A, B)} exists by the axiom schema of comprehension. If K is another set which satisfies x ∈ K iff either (x ∈ A and x B) or (x ∈ B and x A)

, then x ∈ K iff x ∈ C, and so by extensionality K = C.

3.5 a)Given A, B, and C, there is a set P such that x ∈ P iff x = A or x = B or x = C.

b) generalize to four elements

3.5 a) solution: Using A and B along with the axiom of pair provides us with a new set D such that x ∈ D iff x = A or x = B. Now use the axiom of pair again, this time with C alone to obtain the set E such that x ∈ E iff x = C. Now use the axiom of pair one last time, with D and E, to create the set F such that x ∈ F iff x = D or x = E. Finally, apply the axiom of union to F, which yields the set P, where x ∈ P iff x ∈ H for some H ∈ F. Thus, P = {x | x ∈ D or x ∈ E} = {x | x = A or x = B or x = C}.

3.5 b) I'll wait to do this one until someone can verify part a).