Spivak Calculus — Postulate 13 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.

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*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.


*Suppose $S$ is an arbitrary non-empty set of real numbers that is bounded above. Does there exist an algorithm to determine $sup(S)$?
 A: 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
A: 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.
