MEMBERSHIP

ϵ is not reflexive, symmetric or transitive.

xϵx is not a membership statement, it is an identity: x=x.

xϵy →yϵ’x Axiom, 1ϵ{1,2} →{1,2}ϵ’1 (‘ means not)

xϵy & yϵz →’xϵz Axiom, 1ϵ{1} & {1}ϵ{{1},2} but 1ϵ'{{1},2}

IDENTITY: x=x

EQUALITY: x=y, “x” and “y” represent the same thing.

Examples:

Active =A = {on, green,pw,hum,hot}

Def: x=A iff xϵA

xϵA & yϵA → x=y

A={1/3,2/6,3/9,….}

Def: x=A iff xϵA

xϵA & yϵA → x=y

Equality of Sets: A=B: A and B represent the same thing, by Def above.

Standard Definition: every member of A is a member of B and ----.

BUT “is” is not defined, is it identity or equality?

Suppose A ={x,y,z}. if x=w, is w a member of A?

A={blue, green,yellow}

blue =100lb, green =200lb, yellow =300lb

B={100lb, 200lb, 300lb}

Does A=B. By above Def, yes. By Standard Def, depends on what you mean by “is.”

From Suppes AXIOMATIC SET THEORY

Primitive Atomic Formulas (paf): (vϵw), (v=w)

Primitive Formulas (pf), recursive definitions

a) every paf is a pf.

b) if P is a pf, then –P is a pf

c) (P&Q), (PVQ), (P→Q), and (P<̶̶ ̶̶>Q) are pf.

d) (for all v)P, (v exists)P are pf.

Any primitive formula of the form---------------is an axiom. (too difficult to type out, upside down A’s and backward E’s, but irrelevant to the point).

THE POINT: The trouble with the whole scheme is that you are defining truth functions in terms of variables (vϵw), (v=w) which are not defined as truth variables, and that leads to the trouble with

Russells Paradox, derived from the above axiom scheme by using (xϵx) to get:

(xϵx) iff (xϵ’x).

But unless you show (xϵx) is a truth variable with value T (you can’t), Russels paradox is simply a meaningless symbolic axiom and not a Paradox.