Results 1 to 2 of 2

Math Help - infinum supremum

  1. #1
    Newbie
    Joined
    Feb 2010
    Posts
    13

    infinum supremum

    im looking for some help or how to set out this question:

    find the supremum and infinum of the following sets of real numbers, justify answer. state whether they or not they are contained in the set

    X = { 1, \frac {1} {2^2},  \frac {1}{3^2}, \frac {1} {4^2}, ...,  \frac {1}{n^2}, ...}
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,792
    Thanks
    1532
    Quote Originally Posted by murielx View Post
    im looking for some help or how to set out this question:

    find the supremum and infinum of the following sets of real numbers, justify answer. state whether they or not they are contained in the set

    X = { 1, \frac {1} {2^2},  \frac {1}{3^2}, \frac {1} {4^2}, ...,  \frac {1}{n^2}, ...}
    The "supremum" of a set of real numbers is their least upper bound.
    Every number in that decreasing sequence is less than or equal than 1 so it is an upper bound. Since it is in the set itself, no upper bound can be less than 1 so 1 is the least upper bound- the supremum. (If an upper bound for a set is in the set, if it is the largest number in the set, or maximum, then it is the least upper bound= supremum.

    The "infimum" of a set of real numbers is their 'greatest lower bound'.
    There is no smallest number in the set- they just keep getting smaller as you go out the sequenc. Every number in this set is positive so 0 is a lower bound. But it should be clear that the limit of the sequence is 0. If the limit of a sequence is also a lower bound, then it is the greatest lower bound= infimum. That's true because if we claimed that some number, \alpha, greater than the limit, were a lower bound, we could take the " \epsilon" in the definition of limit to be \alpha- L where L is the limit of the sequence. Since there must be numbers in the sequence closer to L than \epsilon, there must be numbers in the sequence less than \alpha- it is NOT a lower bound.

    Although it is not necessary here, if the limit of a sequence is also an upper bound, then it is the least upper bound= supermum.

    For every decreasing sequence, then, the first number in the sequence is the supremum and the limit of the sequence is the infimum.

    For every increasing sequence, the first number in the sequence is the infimum and the limit of the sequence is the supremum.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Finding infinum and supremum of over the quadratic q(x)
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: November 8th 2011, 07:30 AM
  2. Infinum and surpemum
    Posted in the Differential Geometry Forum
    Replies: 5
    Last Post: August 28th 2011, 08:25 AM
  3. supremum ,infinum ,maximum,minimum
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: August 2nd 2011, 04:48 AM
  4. supremum and infinum
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: September 12th 2010, 10:15 AM
  5. Supremum example
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: September 29th 2008, 11:22 PM

Search Tags


/mathhelpforum @mathhelpforum