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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

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    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
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