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Thread: Sets and Orders

  1. #1
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    Sets and Orders

    I. Formalities

    A set S is uncountable if there is no injection from S to a subset of the natural numbers N = {0,1,2,...}

    The Cartesian product A x B is the set of all pairs (a,b) such that a is in A and b is in B. In the case where A=B=R, the set of real numbers, the Cartesian product is the familiar Cartesian coordinate system.

    A relation R on S is a subset of S x S; if (x,y) is in R, write xRy.

    A partial order on S is a relation R on S that satisfies three criteria:

    (1) R is reflexive: For all x in S, xRx
    (2) R is symmetric: For all x,y in S, if xRy then yRx
    (3) R is transitive: For all x,y,z in S, if xRy and yRz, then xRz

    Examples include the "less than or equal to" relation on the real numbers, the "subset" relation on the power set of S, etc.

    A paritally-ordered set is a pair (S,*) where S is a set and * is a partial order on S; where the context is clear, this formalism will be dropped.

    A total order on S is a partial order that satisfies:

    (4) For all x, y in S, either xRy or yRx

    A totally-ordered set is a partially-ordered set whose partial order is a total order.

    A partially-ordered set (S, <=) is locally finite if:

    (5) For all x,y in S, the interval [x,y] = {z in S | x <= z and z <= y} is finite


    II. Problem Statement

    Does there exist an uncountable totally-ordered locally finite set?
    If I am capable of grasping God objectively, I do not believe, but precisely because I cannot do this I must believe. - Soren Kierkegaard
    **** you, I won't do what you tell me

    HOLY CRAP MY BLOG IS AWESOME

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  3. #2
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    Re: Sets and Orders

    Point of clarification about "For all x,y in S, the interval [x,y] = {z in S | x <= z and z <= y} is finite"...


    By this, you do mean that:

    "The interval, [a,b], is finite." <==> "Using the Euclidean metric acting on element a and on element b returns a finite value, Where the Euclidean metric, g: R x R -> R*, defined by g(a,b) = |a - b|."?


    Technically speaking, you need more structure on your set (another relation) to define what it means for a subset to be "finite".
    "Those who can make you believe absurdities, can make you commit atrocities." --Voltaire

  4. #3
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    Re: Sets and Orders

    Quote Originally Posted by GoldPhoenix View Post
    Point of clarification about "For all x,y in S, the interval [x,y] = {z in S | x <= z and z <= y} is finite"...


    By this, you do mean that:

    "The interval, [a,b], is finite." <==> "Using the Euclidean metric acting on element a and on element b returns a finite value, Where the Euclidean metric, g: R x R -> R*, defined by g(a,b) = |a - b|."?


    Technically speaking, you need more structure on your set (another relation) to define what it means for a subset to be "finite".
    If x,y are elements of a poset (S, <=), the interval [x,y] is a set:

    {z in S | x <= z <= y}

    This set is finite iff there exists a bijection from [x,y] to {1,2,...,n} for some n in N, etc. The measure here is cardinality.

    So for example, the interval [0,1] as a subset of R is infinite, since {r in R | 0 <= r <= 1} contains an infinite number of elements.
    If I am capable of grasping God objectively, I do not believe, but precisely because I cannot do this I must believe. - Soren Kierkegaard
    **** you, I won't do what you tell me

    HOLY CRAP MY BLOG IS AWESOME

  5. #4
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    Re: Sets and Orders

    Solution:


    No. Here is the work.
    If I am capable of grasping God objectively, I do not believe, but precisely because I cannot do this I must believe. - Soren Kierkegaard
    **** you, I won't do what you tell me

    HOLY CRAP MY BLOG IS AWESOME

 

 

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