Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Least upper bound of open interval.

  1. Oct 8, 2012 #1
    I am having trouble understanding how there could be a least upper bound for an open interval. If I have (a,b) and i am looking for the least upper bound X which is the number that is less than or equal to the set of Y such that Y> all the numbers in the interval (a,b) when I think about it I can't understand why it would be b.

    If the interval was (1,2) I would first think the lowest upper bound would be just to the left of 2. So 1.9999999 which would be a number with infinite decimals . I don't understand why it would be 2 as my textbooks says.
  2. jcsd
  3. Oct 8, 2012 #2
    See, the real numbers 1.9999999... (with infinitely many decimals; this isn't the same as the 1.9999999 that you wrote!) and 2 are equal. Let me backtrack a little...

    The least upper bound of (1,2), call it s, has to satisfy two conditions:
    (i) s is an upper bound of (1,2), i.e. if 1 < x < 2, then x ≤ s. [This is the upper bound part.]
    (ii) If t is any upper bound of (1,2), then s ≤ t. [This is the least part.]

    So obviously we can't have s < 2, because then there is some number in between s and 2 which is bigger than s, which makes s not an upper bound. E.g. if s = 1.9 we can take 1.95 as our counterexample. Therefore s ≥ 2.

    Now, if s > 2 we can find some number in between s and 2 like before, only this time the number will be smaller than s and bigger than 2. Thus this number will still be an upper bound of (1,2), but it will be less than s, so s will not be a least upper bound. E.g. if s = 2.1 we can take 2.05 as our counterexample. Therefore s ≤ 2.

    Putting these two parts together shows that in fact s = 2. The same reasoning applies to any open interval (a,b), but I chose to stick to (1,2) for concreteness.
  4. Oct 9, 2012 #3

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Filler (JIC):

    For the interval (1,2), if s=2.1, then it would be an upper bound, because there is no number between s and 2 which is less than 2... but it is not the least upper bound because there are still numbers smaller than s which are also upper bounds.

    1.9999... =2 because $$1.999...= 1+9\sum_{n=1}^\infty 10^{-n}$$ ... which is the sum of a geometric series.
    You know how to do those: have a go :)
    Last edited: Oct 9, 2012
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Least upper bound of open interval.
  1. Upper bound/Lower Bound (Replies: 10)