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Math Help - Subdividing an interval

  1. #1
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    Subdividing an interval

    Suppose I have a closed, bounded, non-degenerate interval I = [a, b].
    I have a set X initially containing the points a and b.

    My first action is to find the midpoint of I, call it c, and add c to X.

    I now have two new intervals I_{1}=[a, c] and I_{2}=[c, b].

    My next action is to find the midpoints of I_{1} and I_{2}, lets call them c_{1}, c_{2} and add them to X.

    I now have four intervals.... and I continue this procedure until I can subdivide no more.

    Does X contain all points in the original interval I? Or stated differently, is there a point in I that is not in X?
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  2. #2
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    Re: Subdividing an interval

    Quote Originally Posted by director View Post
    Suppose I have a closed, bounded, non-degenerate interval I = [a, b].
    I have a set X initially containing the points a and b.
    My first action is to find the midpoint of I, call it c, and add c to X.
    I now have two new intervals I_{1}=[a, c] and I_{2}=[c, b].
    My next action is to find the midpoints of I_{1} and I_{2}, lets call them c_{1}, c_{2} and add them to X.
    I now have four intervals.... and I continue this procedure until I can subdivide no more.
    Does X contain all points in the original interval I? Or stated differently, is there a point in I that is not in X?
    What makes you think that "I can subdivide no more?" At each stage your set X is only finite.

    Have you studied the famous Cantor Set ?
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  3. #3
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    Re: Subdividing an interval

    Ignoring the point about "subdivide no more", isn't it obvious that the set of all subintervals does include every point in the original interval? Ever point, p, in [a, b] satisfies a\le p\le b. For any point c, either x\le c or c\le x or both. So if x is in [a, b], either x is in [a, c] or x is in [c,b]. In either case, x is in [a, c]U[c, b]. The general proof can be done by induction on the number of halvings.
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    Re: Subdividing an interval

    Quote Originally Posted by HallsofIvy View Post
    Ignoring the point about "subdivide no more", isn't it obvious that the set of all subintervals does include every point in the original interval? Ever point, p, in [a, b] satisfies a\le p\le b. For any point c, either x\le c or c\le x or both. So if x is in [a, b], either x is in [a, c] or x is in [c,b]. In either case, x is in [a, c]U[c, b]. The general proof can be done by induction on the number of halvings.
    Quote Originally Posted by director View Post
    Suppose I have a closed, bounded, non-degenerate interval I = [a, b].
    I have a set X initially containing the points a and b.

    My first action is to find the midpoint of I, call it c, and add c to X.

    I now have two new intervals I_{1}=[a, c] and I_{2}=[c, b].

    My next action is to find the midpoints of I_{1} and I_{2}, lets call them c_{1}, c_{2} and add them to X.
    At each stage the interval is not added to the set only the new midpoints.

    Then we form a partition of $[a,b]$ with that new set of points and find the midpoints of each new subinterval.

    So after each stage we have a set of a finite collection.
    Thanks from director
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