I don't find Postulate 13 of Spivak's Calculus trivial, nor can I understand why it's true.Postulate 13: Every non-empty set of real numbers that is bounded above has a least upper bound (sup).Why is this postulate true? Any proof/intuition behind it?Edit: Let me pose a few questions.1. Suppose I decide to devise a pathological function R → R which is bounded above but has no supremum. Why will I fail to find such f? If you simply take it as an axiom, there's no guarantee I won't be successful.2. Suppose S is an arbitrary non-empty set of real numbers that is bounded above. Does there exist an algorithm to determine s u p ( S )?

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I don&#039;t find Postulate 13 of Spivak&#039;s Calculus trivial, nor can I understand why it&#039;s true.Postulate 13: Every non-empty set of real numbers that is bounded above has a least upper bound (sup).Why is this postulate true? Any proof/intuition behind it?Edit: Let me pose a few questions.1. Suppose I decide to devise a pathological function       R    →      R   which is bounded above but has no supremum. Why will I fail to find such   f? If you simply take it as an axiom, there&#039;s no guarantee I won&#039;t be successful.2. Suppose   S is an arbitrary non-empty set of real numbers that is bounded above. Does there exist an algorithm to determine   s  u  p  (  S  )?

hoffwnbu · Accepted Answer

It can be based on essential property of real numbers, which is true for any their definition: family of nested closed intervals with length tending to zero have non empty intersection - one point.Proof: let&#039;s consider any set of real numbers   X bounded from above with some number   M i.e.   ∀  x  ∈  X  ,  x  ⩽  M. Take some       x    0    ∈  X and consider interval   [  a  ,  M  ], where   a  &amp;lt;      x    0   and denote it by       σ    0  . Divide       σ    0   in half and denote by       σ    1   right interval, if it contain numbers from   X, otherwise left interval. Continuing in this way we obtain sequence of nested closed intervals with length tending to zero and for each from them there is no members from   X from right. By brought above lemma this sequence necessary have one point in intersection and this point will be exactly   sup for   X.So, as you see, existing of   sup is not trivial or easy question.Of course, the lemma of nested intervals in its own is based on some property, which can be taken as postulate in this case: any increasing sequence bounded from above have limit. Which one of postulate take is different question.

wanaopatays · Accepted Answer

This is the completeness axiom, in other words, the Completeness Axiom guarantees that, for any nonempty set of real numbers   S that is bounded above, a sup exists (in contrast to the max, which may or may not exist (see the examples above).An analogous property holds for inf   S: Any nonempty subset of       R   that is bounded below has a greatest lower bound

I don't find Postulate 13 of Spivak's Calculus trivial, nor can I understand why it's true. Postu

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