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.


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