###### 6/5/12

### Tutorial

Set theory is an extensive topic introduced elsewhere. It can be written as a first order theory.

There is one axiom schema, Abstraction (or Comprehension), which can generate infinitely many axioms

∀y(yε{x:Φ[x]}≡Φ[y])

Axiom Schema of Abstraction (or Specification or Comprehension). The Set Builder Axiom Schema.

And a number of other axioms

(x=y) ≡∀z(zεx≡zεy))

Axiom of Extensionality(x⊂y) ≡∀z(zεx→zεy)

Axiom of Subsets∀x(x∉∅)

Axiom of Empty Set∀x(xεU)

Axiom of Universe Setzε(x∪y) ≡ (zεx∨zεy)

Axiom of Unionzε(x∩y) ≡ (zεx∧zεy)

Axiom of intersectionzε(x-y) ≡ (zεx∧z∉y)

Axiom of Complementzε{x,y} ≡ (z=x∨z=y)

Axiom of Pairzε ℘(x) ≡ (z⊂x)

Axiom of Power Set[These axioms would normally be presented as universally quantified closures ie with leading ∀x∀y etc.]

Then these can be taken as the starting formulas for trees, and the result would be first order (naive) set theory.

Set theory uses many definitions. Trees can accommodate these (but the trees would get a little large and elaborate). We will pass on this.

However there certainly is one theorem worth proving ...

#### Russell's Paradox.

(Naive) set theory is inconsistent, as was discovered by Bertrand Russell. The problem lies with the Schema of Abstraction, then all that is needed is to consider the set of those objects which are not members of themselves. This seems to be a perfectly good defining property for a set (most sets seem to be not members of themselves; for example, the null set ∅ is not a member of ∅ ie ∅∉ ∅). But if we give this set a name, say b, and ask if b is a member of itself: it is iff it isn't (which is a contradiction).

∀y(yε{x:Φ[x]}≡Φ[y])

Axiom Schema of Abstraction (or Comprehension)b={x:x∉y}

So you need to show that

∀y(yε{x:x∉y}≡y∉y), b={x:x∉y} are inconsistent, and that can be done by closing a tree for them

Here's (more than) a hint.

This inconsistency or paradox ruins (naive) set theory.