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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 $f$ : $\mathbb{R}$ -> $\mathbb{R}$ 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 $sup(S)$?

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    $\begingroup$ How does Spivak define the Reals? $\endgroup$
    – LBE
    Jul 31 '20 at 13:45
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    $\begingroup$ You can't prove postulates. As to intuition...well, keep in mind that one "problem" with the rationals is precisely that this does not hold. The set $\{x\in \mathbb Q\}$ such that $x^2<2$ is bounded above but has no (rational) least upper bound. $\endgroup$
    – lulu
    Jul 31 '20 at 13:46
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    $\begingroup$ You need completeness to prove the intermediate value theorem (which is false on $\Bbb Q$). Is this enough "intuition"? $\endgroup$ Jul 31 '20 at 14:33
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    $\begingroup$ @RaiyanChowdhury The IVT is equivalent to Completeness of $\mathbb{R}$. See here: math.stackexchange.com/q/2388577/151611 $\endgroup$
    – ashK
    Jul 31 '20 at 15:30
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    $\begingroup$ In chapter 29, where he defines the real numbers from rationals using Dedekind cuts, you can actually prove all of (P1)-(P13) as theorems. In one of the problems of that chapter he gives an outline of another construction using Cauchy sequences of rationals (by the uniqueness result of the next chapter, these two constructions end up giving you essentially the same thing). $\endgroup$
    – peek-a-boo
    Jul 31 '20 at 16:29
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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's consider any set of real numbers $X$ bounded from above with some number $M$ i.e. $\forall x \in X, x \leqslant M$. Take some $x_0 \in X$ and consider interval $[a, M]$, where $a<x_0$ and denote it by $\sigma_0$. Divide $\sigma_0$ in half and denote by $\sigma_1$ 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.

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  • $\begingroup$ This is a very interesting argument. Provided we can show the distance actually tends to zero. Note positive and decreasing implies convergence by the MCT, but we'd need to show why the distances are converging to zero. $\endgroup$ Jul 31 '20 at 16:23
  • $\begingroup$ Thanks. As to distances, then, they are length of intervals, which we half on each step. $\endgroup$
    – zkutch
    Jul 31 '20 at 17:05
  • $\begingroup$ Yep, this sounds like a good argument to me then. $\endgroup$ Jul 31 '20 at 17:06
  • $\begingroup$ This is essentially replacing one version of completeness (lub property) with another (nested interval principle). $\endgroup$ Aug 1 '20 at 16:06
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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 $\mathbb{R}$ that is bounded below has a greatest lower bound

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  • $\begingroup$ Why do we take it as an axiom if we don't know it's true? Suppose I decide to devise a pathological function $f: \mathbb{R}$ -> $\mathbb{R}$ 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. $\endgroup$ Jul 31 '20 at 15:32

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