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Prove that an upper bound a is the least upper bound

  1. Mar 2, 2014 #1
    1. The problem statement, all variables and given/known data
    Let A be a non-empty subset of R (real numbers) and a an upper bound in R for A. Suppose that every open interval I containing a intersects A (so the intersection is non-empty). Show that a is a least upper bound for A.

    3. The attempt at a solution

    I've seen the prettier proof which involved a contradiction, but just wanted to know if my rather non-rigorous argument makes sense, and if it can be made more precise:

    Let b=sup(A). Let an interval I=(x,y) satisfy the above (contains a, intersects A). If we take the limit as y approaches b, then a must approach b=sup(A) and so since the limit of this I is just another open interval, and a=sup(A) in this limit, then a=sup(A).
     
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  3. Mar 2, 2014 #2

    Dick

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    I would suggest you stick with the pretty proof. 'a' is supposed to be a fixed upper bound. Talking about it 'approaching' something really doesn't make any sense.
     
  4. Mar 2, 2014 #3
    I don't think you can take the limit as y approaches b because how do you know that (x, y) would still contain a? what if b < a and then as you take y closer and closer to b, there's a point where the interval doesn't contain a anymore.
     
  5. Mar 3, 2014 #4
    Yeah I guess I assumed a to be in the interval not matter how you took I, as long as it intersected with A (so not some fixed a).

    Anyway, is this correct? I didn't write down it down but I remember the idea went like this:
     
  6. Mar 3, 2014 #5

    micromass

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    That's the right idea, but I'd want to see how you construct the interval I (or why it exists).
     
  7. Mar 3, 2014 #6

    Dick

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    And I'd want you to say exactly why the existence of the open interval contradicts a being a least upper bound.
     
  8. Mar 4, 2014 #7
    @micromass: Hmm...I'm not sure of a rigorous justification, but I would say because if there are an 'infinite' number of real numbers between a and sup(A) if a is not sup(A), it's possible to construct an open interval around it. *

    @Dick: Contradicts a being a least upper bound? Do you mean a not being a least upper bound? It contradicts the assumption because we've assumed that every open interval I containing also intersects with A. Assuming a is not sup(A), then the construction of an open interval I around it such that I does not intersect with A leads to a contradiction, therefore a must be sup(A).

    *I've also been thinking about how to show that if a=sup(A), then it's not possible to have an open interval I containing a, NOT intersect with A (i.e. if it contains a=sup(A), it must intersect with A).
    I think the case if a is in A is 'trivial' as they say, but I'm unsure as to how to make rigorous the argument for the case where a is not in A.
     
  9. Mar 5, 2014 #8

    micromass

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    If ##\textrm{sup}(A)<a##, then think about an interval centered at ##a## with radius ##(a-\textrm{sup}(A))/2##.

    That's not really needed for the proof here. But you should proceed by contradiction. Assume there is an open interval ##(a-\varepsilon,a+\varepsilon)## that contains no elements of ##A##. Prove that ##a-\varepsilon## is an upper bound.
     
  10. Mar 8, 2014 #9
    Hmm...here's an attempt:

    If [itex]a>sup(A)[/itex], then [itex]a-sup(A)>0[/itex]. Let the the interval I be, for some ε>0 [itex]I=(a-\frac{a-sup(A)}{2}, a+\epsilon)=(\frac{a+sup(A)}{2} a+\epsilon[/itex].

    Yeah, it was just out of curiosity :p
    Still working on it though, I think I've got the idea but just refining it.
     
  11. Mar 8, 2014 #10

    micromass

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    It's probably obvious to you, but you also need to prove this doesn't intersect ##A##.
     
  12. Mar 17, 2014 #11
    Sorry for the delayed response, had - still have lots assessable work to do (procrastinating on that!).

    Anyway, is it correct to say that since [itex]inf(I)=\frac{a+sup(A)}{2}>sup(A)[/itex], then I doesn't intersect with A?


    And for the other part (show if a=sup(A) is in an open interval I, I must intersect with A).

    Suppose a=sup(A) and that there exists an open interval I containing sup(A) that does not intersect with A. Let I=(sup(A)-ε, sup(A)+ε). Since the intersection of I and A is the null set, this implies sup(A)-ε is an upper bound for A which contradicts a=sup(A).

    Or I think I could explain further and get a different contradiction: I not intersecting A implies sup(A)-ε is an upperbound because if it weren't, then there would be some element from A greater than sup(A)-ε and less than sup(A), i.e. in I, which would contradict the assumption that I does not intersect A.
     
  13. Mar 17, 2014 #12

    micromass

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    Seems ok to me.
     
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