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Math Help - Set Membership, Equality, and Russells Paradox

  1. #1
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    Set Membership, Equality, and Russells Paradox

    MEMBERSHIP
    ϵ is not reflexive, symmetric or transitive.
    xϵx is not a membership statement, it is an identity: x=x.
    xϵy →yϵx Axiom, 1ϵ{1,2} →{1,2}ϵ1 ( 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 As and backward Es, 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 (xϵx).

    But unless you show (xϵx) is a truth variable with value T (you cant), Russels paradox is simply a meaningless symbolic axiom and not a Paradox.
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  2. #2
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    Re: Set Membership, Equality, and Russells Paradox

    Quote Originally Posted by Hartlw View Post
    MEMBERSHIP
    ϵ is not reflexive, symmetric or transitive.
    xϵx is not a membership statement, it is an identity: x=x.
    I have no clue what you mean by this. Grammatically you are saying that " x\in x" is the same as " x= x" but surely you don't mean that. What is true, is that " x\in x" is never true.

    xϵy →yϵx Axiom, 1ϵ{1,2} →{1,2}ϵ1 ( 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?
    We can't redefine every English word for mathematics. How are you distinguishing between "identity" and "equality". Mathematically, two objects are identical if and only if they are equal. Where have you ever seen a different definition for either "identical" or "equal"?
    (We do, in some cases, give different definitions for "equivalent" in different fields of mathematics. The general category of "equivalence relations" includes "equal" (or "identity") but also other things.)

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

    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.
    By the 'above Def.", yes, but "blue= 100 lb" is a very strange definition! Assuming that "blue" and "100 lb", etc. are simply labels for the same thing and have nothing to do with our usual definitions of those words, then, yes, A= B. Now, what do you mean by the "standard definition"? The "standard definition" of what? What you give is the "standard definition" of "equals" (and "are indentical) for sets. If you mean "the standard definition(s)" of "blue", "200 lb", etc. then A\ne B, but then "blue= 200 lb" is no longer true.

    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 As and backward Es, 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 (xϵx).

    But unless you show (xϵx) is a truth variable with value T (you cant), Russels paradox is simply a meaningless symbolic axiom and not a Paradox.[/QUOTE]
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  3. #3
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    Re: Set Membership, Equality, and Russells Paradox

    The number 300 on a force gauge means 300lbs, but they are not the same. But 300 on the force gauge = 300lbs force because both sides represent the same thing, an actual physical force of 300lb. The force gauge can have colors instead of numbers.

    The only reason I posted this thread is that the books I have on axiomatic set theory, let alone Set Theory, do not define ϵ or =. Without a definition of these things, how can you even talk about Russels Paradox, ar anything else for that matter.
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