Results 1 to 8 of 8

Math Help - Sequence and supremum

  1. #1
    Newbie
    Joined
    Oct 2009
    Posts
    23

    Sequence and supremum

    Suppose that S is a nonempty set of real numbers and that theta is an upper bound of S. Prove that the following conditions are equivalent:
    (1) We have theta = Sup S.
    (2) There exists a sequence (X sub n) in S such that (X sub n) approaches theta as n approaches infinity.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by Slazenger3 View Post
    Suppose that S is a nonempty set of real numbers and that theta is an upper bound of S. Prove that the following conditions are equivalent:
    (1) We have theta = Sup S.
    (2) There exists a sequence (X sub n) in S such that (X sub n) approaches theta as n approaches infinity.
    I disagree.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Oct 2009
    Posts
    195
    Quote Originally Posted by Drexel28 View Post
    I disagree.
    I agree with Drexel. If you take a bounded sequence with a maximum and an infimum, then (1) does not imply (2).

    In other words, if you choose a sequence such that you'll ever get sufficiently close ( < \varepsilon ) to \theta , the sequence cannot continue on until infinity. Or, if you choose a sequence that will continue on until infinity, it will always reside in the \varepsilon -neighborhood of L.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    2 does not imply one. Consider [0,1]=S and x_n=\frac{1}{2}. Then \left\{x_n:n\in\mathbb{N}\right\}\subseteq S and x_n\to\frac{1}{2} but surely \frac{1}{2}\ne\sup\text{ }S
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Oct 2009
    Posts
    195
    Quote Originally Posted by Drexel28 View Post
    2 does not imply one. Consider [0,1]=S and x_n=\frac{1}{2}. Then \left\{x_n:n\in\mathbb{N}\right\}\subseteq S and x_n\to\frac{1}{2} but surely \frac{1}{2}\ne\sup\text{ }S
    There exists, not for all. This counterexample is false.

    That is, there does exist a sequence in S, namely (x_n) = 1 - \frac{1}{n}, such that (x_n) \to \theta . Now we must either prove that for any choice of S there exists a sequence like this, or that there exists an S for which no (x_n) exists that implies (1)
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by davismj View Post
    There exists, not for all. This counterexample is false.
    What? 2 says that if I can find a sequence in S whose limit is \theta then \theta=\sup\text{ }S. How did my counterexample not work?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Oct 2009
    Posts
    195
    Quote Originally Posted by Drexel28 View Post
    What? 2 says that if I can find a sequence in S whose limit is \theta then \theta=\sup\text{ }S. How did my counterexample not work?
    Yea, you're right. Somehow, I don't think that's what they meant, though.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by davismj View Post
    Yea, you're right. Somehow, I don't think that's what they meant, though.
    Of course it's not what they meant. That doesn't mean we bear the burden of interpreting there obviously typo.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Sequence, Supremum, Infimum
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: March 12th 2010, 05:26 AM
  2. supremum
    Posted in the Differential Geometry Forum
    Replies: 5
    Last Post: March 1st 2010, 04:05 AM
  3. Supremum
    Posted in the Calculus Forum
    Replies: 3
    Last Post: October 22nd 2008, 12:29 PM
  4. supremum
    Posted in the Calculus Forum
    Replies: 1
    Last Post: August 9th 2008, 11:02 PM
  5. Supremum
    Posted in the Calculus Forum
    Replies: 5
    Last Post: September 20th 2006, 08:39 AM

Search Tags


/mathhelpforum @mathhelpforum