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Supremum = least upper bound, anything > supremum?

  1. Feb 23, 2016 #1
    The supremum is defined as the "LEAST" upper bound. The word "least" makes me think, there is a "MOST" upper bound, or at least something bigger than a "least" upper bound.

    For a set of numbers, is there anything larger than a supremum? Supremum is analogous to a maximum, but I don't understand what it's called "least" upper bound.
     
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  3. Feb 23, 2016 #2

    PAllen

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    Companion to least upper bound is greatest lower bound.

    LUB is not like a maximum. It is the minimum of upper bounds.
     
  4. Feb 23, 2016 #3
    Thanks, Could you illustrate this with a set of numbers? I still do not understand.
     
  5. Feb 23, 2016 #4

    PAllen

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    Sure. Consider the set of numbers defined by decimal fractions beginning with a decimal point. Any number greater than or equal 1 is an upper bound. The least upper bound is 1.
     
  6. Feb 23, 2016 #5
    Oh I think I understand. In this example, we considered only the set of numbers beginning with a decimal point.
    If we consider all real numbers, is the least upper bound = upper bound = ##\infty##?
     
  7. Feb 23, 2016 #6

    PAllen

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    Infinity isn't a real number. So there is no LUB for all real number among the real numbers. If you consider the reals as a subset of an extended real number set, then you can make this statement true.
     
  8. Feb 24, 2016 #7

    Erland

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    Consider the open interval I = ]0,1[. Then, for example, that x ≤ 2, for all x ∈ I. It is also true that x ≤ 5, for all x ∈ I. This means, by definition, that 2 and 5 are upper bounds of I. A set which has (at least) one upper bound is said to be bounded above. Thus, I is bounded above, while for example [0,∞[ is not: the latter has no upper bound.
    There, are, of course, infinitely many upper bounds of I. Indeed, any number u ≥ 1 is an upper bound of I. The set of all upper bounds of I is then the interval [1,∞[. This set has a smallest (least) element: s=1, which is then the least upper bound of I. This least upper bound s of I is also called the supremum of I.

    The axiom of a least upper bound says that every set of real numbers which is bounded above (i.e. it has an upper bound) has a least upper bound, or a supremum. If the set is not bounded above, then it has no upper bound at all, and hence no least upper bound.
     
  9. Feb 24, 2016 #8
    For the purpose of least upper bound i.e., the sup (and greatest lower bound, i.e. the inf) of any set of numbers, it's more convenient, and standard in math, to include the possibility of either +∞ or -∞ (the plus sign is optional), if one of them is the appropriate answer. It's less convenient — and less informative — to have to say that the sup or inf is "undefined".

    This way the sup and inf of an arbitrary set of real numbers is uniquely defined.

    So Yes, the sup of the set of all real numbers is ∞, and its inf is -∞.

    It's good to note that the sup or inf might or might not belong to that set.

    (((It will probably be surprising to learn that the sup of the empty set is normally taken to be -∞, just as the inf of the empty set is normally taken to be +∞.)))
     
  10. Feb 24, 2016 #9

    HallsofIvy

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    The set {0, 1, 2, 3, 4} has every number larger than or equal to 4 as upper bound. The least upper bound is 4.

    The set [0, 1], the set of all real numbers from 0 to 1 including both 0 and 1 has all real numbers larger than or equal to 1 as upper bounds. The least upper bound is 1.

    The set (0, 1), the set of all real numbers from 0 to 1 but not including 0 and 1 still has all real numbers larger than or equal to 1 as upper bounds. The least upper bound is 1.

    If an upper bound of a set is also in the set, as in the first two examples above, the upper bound is the least upper bound and is also a "maximum" for the set.

    In the third example above, the set has NO maximum, or largest member, so the least upper bound is not in the set.
     
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