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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 $sup(S)$?

Raiyan Chowdhury
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    How does Spivak define the Reals? – LBE Jul 31 '20 at 13:45
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    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. – lulu Jul 31 '20 at 13:46
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    You need completeness to prove the intermediate value theorem (which is false on $\Bbb Q$). Is this enough "intuition"? – Angina Seng Jul 31 '20 at 14:33
  • @lulu Of course you cannot prove postulates. However, only statements which are absolutely trivial should be taken as axioms. In other words, it should be clear to the reader or listener that every axiom assumed is true, otherwise the validity of all your arguments are compromised. Note this extends beyond mathematics. If your assumptions are false, why should I believe your argument? – Raiyan Chowdhury Jul 31 '20 at 15:12
  • @user587399 The reals are defined as sets of rational numbers. – Raiyan Chowdhury Jul 31 '20 at 15:16
  • @AnginaSeng Unfortunately, no. The IVT is used for functions which are continuous on an interval $I$. A set can be any arbitrary collection of real numbers. Even assuming the IVT you cannot reason the postulate. – Raiyan Chowdhury Jul 31 '20 at 15:23
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    @RaiyanChowdhury The IVT is equivalent to Completeness of $\mathbb{R}$. See here: https://math.stackexchange.com/q/2388577/151611 – ashK Jul 31 '20 at 15:30
  • If you follow the (there are several ways to do this) construction of the reals from the rationals, it will help you to see how the postulate dovetails with "intuition." – Matematleta Jul 31 '20 at 15:31
  • How could a statement like this possible be "clear to the reader or listener"? The construction of the real line is very far from trivial. Of course, the whole idea behind the reals was to "fill in the holes in the rationals" so it would be very disappointing were this property not to hold. But, of course, the expectation must then be that one shows that some natural construction or other satisfies this axiom. Again, though, there is no easy or obvious construction. – lulu Jul 31 '20 at 15:40
  • @AshishK Why is this postulate logically equivalent to the completeness of $\mathbb{R}$? If you were to cut the real number line at a point, and throw all the numbers to the left into a set, it is clear the set has a supremum. Either it's the maximum, if it exists, otherwise it's the limit point. However, this postulate is making a much more general claim for any arbitrary set, one which is bounded above. – Raiyan Chowdhury Jul 31 '20 at 16:00
  • @lulu "it would be very disappointing were this property not to hold". Agreed, so it is an enormous leap of faith we're taking in assuming the validity of this statement. Which is of course why it's called an axiom. Which means no one knows if it's actually true. But it's an axiom, and all axioms have to be true, right? – Raiyan Chowdhury Jul 31 '20 at 16:13
  • I don't really need to see a "proof" per se. Just need to know that someone knows why it's true. This is a theorem like any other theorem. "There any infinitely many primes." This could have easily become a postulate had Euclid not provided us with a proof. – Raiyan Chowdhury Jul 31 '20 at 16:17
  • Spivak takes P13 as a postulate, because it is something which is false in $\Bbb{Q}$, so if $\Bbb{R}$ is supposed to "fill in the gaps of the rationals" then it had better satisfy an additional property like this one. Also, from the way he structured the book, it's clear that he doesn't want to overload the reader with too many abstract ideas too early (and having studied from Spivak, I'm glad he didn't). If you're eager then of course you can go right ahead to Chapter 29 where he gives a construction of $\Bbb{R}$ from $\Bbb{Q}$ and shows (in Chapter 30) that it is essentially unique. – peek-a-boo Jul 31 '20 at 16:21
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    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). – peek-a-boo Jul 31 '20 at 16:29
  • @peek-a-boo See my pathological function problem. To me, I think this postulate is remarkable, and deserves a brilliant argument. I've read the book a few times but now I'm going through with a fine tooth-comb. Too many readers seem to not look past the surface when dealing with statements they don't totally understand. – Raiyan Chowdhury Jul 31 '20 at 16:30
  • The issue is that at the stage where he introduces P13, Spivak has not yet defined what $\Bbb{R}$ is. Really all he is saying is "Let us assume the existence of a set $\Bbb{R}$ along with operations $+,\cdot, <$ which satisfy P1-P13". Then everything he does from this point relies heavily on the assumption of existence. This is really a leap of faith which he is making, but only a temporary one, because he resolves it at the end of the book. So, regarding your pathological function, yes you will fail to find such an $f$ because by construction of $\Bbb{R}$, it has the supremum property. – peek-a-boo Jul 31 '20 at 16:45
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    Or said differently, let us assume the existence of a set $H$ and operations $+,\cdot,<$ which satisfies P1-P13 (i.e a complete ordered field). Then of course because $H$ satisfies P13 (by assumption), it will be true that every function $f:H \to H$ which is bounded above will be such that its image has a supremum in $H$. This is simply true because you're assuming the existence of $H$ which satisfies all those properties. Now, it could have been the case that such an $H$ doesn't exist; and this would be a sad situation indeed, but turns out such an $H$ does exist, and we call it $\Bbb{R}$. – peek-a-boo Jul 31 '20 at 16:53
  • I think you are confused. Prior to the constructions, people got by on hope. It seemed likely enough that a set with this sort of properties existed and using those properties enabled them to prove exciting and powerful theorems. Of course, there was the risk that someone could prove that these axioms were inconsistent. That was resolved by constructing a set with those properties. These days, of course, we can use Cauchy sequences or Dedekind sets to create systems which we can prove satisfy these postulates, and we can then call these systems the real numbers. – lulu Jul 31 '20 at 16:59
  • @lulu Edit: Sorry I missed your last point. Yep, if it's proven to satisfy the postulates that works. – Raiyan Chowdhury Jul 31 '20 at 17:10
  • But...I assure you that you will not find those constructions "intuitive". They really, really aren't. They do get the job done. That is, they establish the consistency of the postulate system. – lulu Jul 31 '20 at 17:16
  • @peek-a-boo Great answer. I'd assume $\mathbb{R}$ is one such set satisfying the postulates, but there are infinitely many. Categorically differentiating $\mathbb{R}$ from other sets satisfying the postulates would be another topic. So I guess this is quite clear, what we've done is come up with a list of postulates, then we've proven $\mathbb{R}$ satisfies each of these. But I find it very strange in that case for the postulates to be the defining property of the real numbers, but I guess Spivak clarifies this later in the book. – Raiyan Chowdhury Jul 31 '20 at 20:32
  • @RaiyanChowdhury You can have many different sets which satisfy all these properties, but this construction is unique in the sense that if $(H_1,+1,\cdot_1, <{1})$ and $(H_2, +_2, \cdot_2, <_2)$ are two such complete ordered fields (i.e satisfy P1-P13), then there is a ordered-field isomorphism $f:H_1\to H_2$. This is the uniqueness part discussed in Chapter 30. – peek-a-boo Jul 31 '20 at 20:46
  • Essentially what this is saying is that while you may have different set-theoretic descriptions for the real numbers, they're all trying to describe the "same thing". This is a reinforcement of something which my professor drilled to me: "it usually doesn't really matter 'what' an object is, but more what properties it has". As lulu mentions above, you probably won't find the construction very illuminating, because for it to make sense, you need some intuition on what it is we're trying to do. It's just one of those things we do to make sure we haven't introduced any contradictions. – peek-a-boo Jul 31 '20 at 20:46
  • It will best if you can try to understand that this property does not hold for system of rationals and that's the sole reason for inventing reals. Unfortunately people fixed this deficiency of rationals by assuming this property of reals and much later tried to define reals in terms of rationals so that the property is no longer an axiom but rather a theorem. I prefer the approach which does not assume this as an axiom. – Paramanand Singh Aug 01 '20 at 16:09
  • @ParamanandSingh Agreed. As stated the postulate appears quite convoluted. As a reader, I would have preferred the axiom to be the completeness of $\mathbb{R}$, and this postulate to be proven as a consequence. It does not seem trivial how completeness is equivalent to this postulate. – Raiyan Chowdhury Aug 01 '20 at 23:23
  • I suggest you look at this answer which describes various ways to express completeness of $\mathbb{R} $. – Paramanand Singh Aug 01 '20 at 23:34

2 Answers2

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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

  • 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. – Raiyan Chowdhury Jul 31 '20 at 15:32
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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.

zkutch
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