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Real Analysis- least upper bound and convergence

  1. Sep 11, 2007 #1
    I'm having a little difficulty understanding Epsilon in the definition of convergence. From what the book says it is any small real number greater than zero (as small as you can imagine???). Also, since I don't quite grasp what this epsilon is and how it helps define convergence, I am having difficulty applying it to the following problem:

    Let b=Least upper bound of a set S (S is a subset of the real numbers) that is bounded and non empty. Then Given epsilon greater than 0, there exists an s in S such that (b-Epsilon)<= s <= b.

    I started by proving that there exists an s in S, but I cannot figure out how to relate this all to epsilon.
  2. jcsd
  3. Sep 11, 2007 #2
    I understand the definitions of cauchy sequences and the cauchy criterion, but I don't see how I can relate them to this problem. I think what I am confused with is what kind of S we are dealing with here. FOr example, doesn't S={1*, b} satisfy these conditions but i don't see how i can use the cauchy sequence etc with the subset...
  4. Sep 12, 2007 #3
    You know that b is the least upper bound. So b-1, for example, isn't an upper bound. Neither is b-1/2, or b-1/4, etc... b-epsilon will not be an upper bound for any choice of epsilon>0.

    Now, since b-epsilon is not an upper bound, there must be an s in S such that b-epsilon<=s, or else b-epsilon would be a lesser upper bound.

    The set S is arbitrary (well, it has to be nonempty if b is not to be [itex]-\infty[/itex]). I don't know what you mean by 1* (it's the real number 1, defined in some way from the rational number 1?), but I assume you mean 1*<b. Then S has the the least upper bound b. That's right. But note that S will not always contain b. For example, the set {-1,-1/2,-1/3,-1/4,...} has the least upper bound 0, but 0 is not in the set.

    Cauchy sequences don't enter into this problem, don't worry about them.

    Does that answer your questions? :)
    Last edited: Sep 12, 2007
  5. Sep 12, 2007 #4


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    The point of convergence of a sequence is that the numbers in the sequence get "closer and closer" to the limit. The purpose of [itex]\epsilon[/itex] is to quantify, make clear, what "closer and closer" means. Saying "Given [itex]\epsilon[/itex]> 0 , there exist N such that if n> N, then |an- L|< [itex]\epsilon[/itex]" means that we can get as close as we please (< [itex]\epsilon[/itex]) to L just by going far enough on the sequence (n> N).

    Now, your problem here is only indirectly related to convergence. The assertion is "Given [itex]\epsilon> 0[/itex], there exist some s in S such that b-[itex]\epsilon< s \le b[/itex]". Since b is an upper bound on S, there are no members of S larger than b: that gives you the "[itex]s \le b[/itex]" part. Suppose there were NO members of s larger than [itex]b-\epsilon[/itex]. What does that make [itex]b-\epsilon[/itex] and how does that contradict the fact that b is the least upper bound?
  6. Sep 13, 2007 #5
    Thank you so much this helps a lot. I guess that the proof would be easiest assuming there is no such s and then show that b-epsilon is thus greater than all s thus giving a contradiction to b being the least upper bound. This problem seems almost trivial now i know how to tackle it. :)
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