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Bisecting indefinitely a real interval

  1. Jan 13, 2006 #1
    (A, B) is an interval in the real line. If I take the middle point (B-A)/2, and then take the middle point of each of the two subintervals of (A,B), and then take the middle point of each of the four subintervals of (A, B) and I go on this indefinitely, can I be sure that sooner or later I will touch any point of (A, B)?

    (I know that there are infintely many irrationals in (A, B) but I am carrying this process "to the limit").

    Maybe dumb question but I need your support.

    Thanks.
     
  2. jcsd
  3. Jan 13, 2006 #2

    TD

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    I'm not sure if you mean it like that but when you write the interval (a,b), that is an open interval meaning the end points are excluded (a and b itself). Of course, you'll never reach a and b this way.
    The interval with the endpoints included, which is the closed interval, is usually written [a,b].
     
  4. Jan 13, 2006 #3

    matt grime

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    What do you precisely mean by 'touch' and 'go on indefinintely'?

    I think the answer you want is 'yes, you can get arbitrarily close to any number in the interval by considering the points in this construction: given we've done the split n times then two neighbouring points we construct are 2^n apart. Any number in the whole interval lies in one of these subintervals so there is some point we construct that is less than 2^{n+1} from it, and we can make 2^{n+1} as small as we choose.
     
  5. Jan 13, 2006 #4
    bisecting interval

    Why can't I put a space between paragraphs?

    TD, I mean [A, B].
    Thanks, Matt. I think your answer is enough.
    The reason of my question was this. (Excuse but I can't use latex. Any time i put the symbols i got: "Latex graphic is being generated. Reload this page in a moment", and never go beyond. So I have to put all in words).
    With "P" I mean any partition of [a, b] and with "P_n" I mean a partition of [a,b] constructed bisecting n times the interval. So any P_n is a P but not viceversa.
    In one proof Apostol implies that, as the Riemann integral of the function "f" over [a, b] is equal to the supremum of {lower sums of "f" over any partition P of [a, b]}, then it is also equal to the supremum of {lower sums of "f" over any partition P_n of [a,b]}.
    My doubt was: The set {lower sums of "f" over a partition P_n of [a, b]} is a proper subset of the set {lower sums of "f" over a partition P of [a, b]}, so it is not automatically valid to say that the supremum of the bigger set is also the supremum of the smaller one. There could be an element in the bigger set that: 1) is less than the supremum of the bigger set, and 2) it is not on the smaller set. So this element would be an upper bound of the smaller set and less than the supremum of the bigger set. So the supremum of the bigger set would not be the least upper bound of the smaller set.

    But your answer tells me that if I take any partition P I can aproximate beoynd all limit it by some partition P_n so that element I mentioned can not exist and therefore the supremum of the bigger set is also supremum of the smaller one.
    Excuse the poor redaction. I am not so fluent in english.
     
  6. Jan 13, 2006 #5
    This problem can be simplified.

    Imagine, you want to get to the number 2. You start by adding the half of 2, 1. 0+1=1, *dUh*. Then, you add half of 1, 0.5. You get 1.5. Next, you get 1.75, 1.875, etc, etc. If you go on infinitely, you will eventually get to 2.
     
  7. Jan 13, 2006 #6

    matt grime

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    Ah, the criterion of integrability. Yes, it is slightly surprising that its true: if you can find one sequnce of partitions whose 'length' tends to zero and for which the upper and lower riemann sums converge and agree then the function is integrable (ie works for all partitions when the length tends to zero).

    I think the 'big' reason for this is that since the real numbers are defined in terms of sequences you can prove things in terms of sequences. Some spaces are far too complicated for you to determine behaviour based upon merely sequences (things indexed by N), and you need to pass to things called ultra-filters.
     
  8. Jan 13, 2006 #7
    Thanks, Matt.

    Castilla.
     
  9. Jan 13, 2006 #8
    Thanks, nagsjunk, your example is fine though it is not exactly the case.

    Suppose you have interval [0, 5]. You want to "touch" 2. You can not bisect [0, 2] and then bisect [1, 2] and then bisect [1.5, 2], that way is not allowed. The only "road" you are allowed to take is to bisect [0, 5] and then to bisect [0, 2.5], and then to bisect [1.25, 2.5], and then to bisect etcétera...
     
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