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Help with nested intervals in Courant please

  1. Aug 10, 2010 #1
    I'm having trouble grasping this concept. This is the part in question:

    Why is the point x uniquely determined by the nested sequence? if i pick two rational numbers, no matter how close together they are, surrounding say the square root of two, shouldn't there always be another two rational numbers that are closer?

    What does he mean when he says the distance between x and y would exceed the length of In? shouldn't you be able to decrease the lengths of both of them as much as you need to because there is an infinite amount of points?

    What does he mean by a preassigned positive number? I'm sure that's the reason why I'm not understanding this... you have to place some sort of cap on how small you can go, right?

    He says in the footnotes:

    Why does this matter? Again, couldn't you just take the point on L directly in the middle of the two open endpoints and have an x in the middle of the smallest In?

    Please help this is driving me crazy...
  2. jcsd
  3. Aug 11, 2010 #2
    Can anybody help me please?
  4. Aug 12, 2010 #3


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    That last point, that the intervals must be closed, is important because we are talking about an infinite sequence of intervals. Consider the case with In= (0, 1/n). That is, the intervals are (0, 1), (0, 1/2), (0, 1/3), (0, 1/4), ...

    Every one of those intervals is non-empty. What about their intersection? Since none of the intervals contains a negative number or 0, their intersection contains no negative number and does not contain 0. Let "x" be a positive real number. The 1/x is also a positive real number and by the Archimedian property, there exist a positive integer, N, such that N> 1/x. Since x and N are positive, then, x> 1/N and x is NOT in IN= (0, 1/N). That shows that no positive number can be in the intersection of those intervals and so the intersection is empty.

    You have a series of nested intervals, I1, I2, ..., In, ... with each interval a subset of the previous one and length(In) going to 0. We can write each interval in the form [itex]I_n= [a_n, b_n][/itex]. Because the intervals are nested we have [itex]a_n\le a_m\le b_m\le a_m[/itex] for n< m. That says that the sequence [itex]\{a_n\}[/itex] is an increasing sequence having every [itex]b_k[/itex] as an upper bound. By the "monotone convergence" property, that sequence converges to some limit- call it "a". Of course, that a is also the least upper bound on the set [itex]\{a_m\}[/itex] and, since every bn is an upper bound, we have [itex]a\le b_n[/itex]. That is, [itex]\{b_n\}[/itex] is a decreasing sequence having a as a lower bound. By "monotone convergence" again, that sequence converges to some number, b. Since b is the greatest lower bound for the sequence and a is a lower bound, [itexs]a\le b[/itex]. Every number in [a, b] is less than or equal to every bn and larger than or equal to every an and so in all of the intervals. The intersection of all intervals is [a, b] which is non-empty (if a= b then it contains the single point a- if a< b, then it contains an infinite number of points).

    Now suppose a< b- that is suppose the interval [a, b] contains more than one point. Let "x" and "y" be two points in that interval. Notice that we have chosen x and y at this point- they are not "variables" now. They have a specific positive distance between them. Yes, given any N, we could choose x and y so that the distance between x and y is less than bn- an[/b] but that is not what we are doing here. That would be picking N first, then picking x and y. Here we are picking x and y first. Now, once the two points x and y are fixed, with y not equal to x, the distance between x and y, |x- y| is a positive number. Since the lengths bn- an go to 0, there exist N such that bn- an< |x- y|- and so it is impossible for both x and y to "fit" inside that interval and so cannot be in the intersection of all the intervals. That shows there cannot be two distinct points in the interval [a, b]. We must have a= b and the intersection of those nested, closed, intervals is a single point.
  5. Aug 13, 2010 #4
    thank you! i understand now, I'm gonna go reread that part in the book now, should make a lot more sense this time around.
  6. Aug 25, 2010 #5
    Nope. I still dont get it.

    If a = b, since a is increasing and b is decreasing and [a,b] is the intersection of all I sub n (sorry im on my cell phone computer isnt working), that means that and b must be rational, right? since all the endpoints of the intervals that we chose had to be rational. so how do you describe an irrational number then with closed intervals?

    sorry for being so thick headeim just not understanding it its been bugging me for weeks now
  7. Aug 25, 2010 #6


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    [3, 4]
    [3.1, 3.2]
    [3.14, 3.15]
    [3.141, 3.142]
    [3.1415, 3.1416]
    [3.14159, 3.1416]
    [3.141592, 3.141593]
    . . .
  8. Aug 26, 2010 #7


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    Maybe you should think in terms of convergent sequences. The end points of the intervals are just respectively increasing and decreasing sequences of rational numbers converging to a common irrational number.
  9. Aug 26, 2010 #8
    That would make sense except he just showed that there is no number in a series of infinate converging intervals so what is the difference if you use closed intervals? The only difference is you are including the end points. so if there was no number contained in all open I sub n then the number contained in all closed I sub n must be one of the end points, right? And we were only allowed to chose rational endpoints.

    so my question is how do you enclose an irrational number in all closed I sub n with rational endpoints if you cant enclose an irrational number in all open I sub n with rational endpoints?
  10. Aug 27, 2010 #9


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    No, what was shows is that there exists a series of converging intervals which have no numbes in their intersection. There are certainly converging intervals (open or closed, take your pick) that contain one or more numbers.

    I gave an example above.
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