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Source of background information:《The Real Analysis Lifesaver》ISBN:9780691172934

P37: “the axiom of completeness”—here, completeness is just another word for the least upper bound/greatest lower bound P32:Definition 4.10. An ordered set S has the least upper bound property if, for every nonempty subset E which is bounded above in S, sup E exists in S.

------All my questions stem from the proof process of Theorem 4.1 in this book.P27

Summary of the theorem's proof:Let A be the set of all $$ p \in \mathbb{Q} $$ such that $$p^2<2$$ and p > 0. To show that A has no single largest number, we must show that for any element of A, there is a greater element also in A


The above is from the book, and the following are my questions.

---------------"Subsequent discussions explain that Q (the set of rational numbers) does not possess the least upper bound property. In other words, if we consider only sets within the set Q, there are some sets for which we cannot find an upper bound! Does this lead to the situation where the limits of certain sequences of rational numbers cannot be found within the set of rational numbers? Therefore, can the statement above be summarized as 'Is completeness of real numbers understood as closure under limits in the real number system?' Is the incompleteness of rational numbers equivalent to the lack of closure under limits? For example, consider constructing a sequence using a series. As n approaches infinity, the limit of this sequence results in an irrational number. Does this imply that rational numbers are incomplete?"

$$ \frac{1}{1^1}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2} \ldots \cdots \cdots \cdots=\frac{\pi^2}{6} $$

Can the principles of the least upper bound and the Cauchy convergence criterion be unified based on the above understanding?

Using the least upper bound principle to determine the completeness of real numbers or the incompleteness of rational numbers seems quite obvious and practical. Rational numbers lack the property of having a least upper bound, hence they are incomplete.

However, it seems that the statement process of the Cauchy convergence criterion cannot be directly applied to determine the completeness of real numbers or the incompleteness of rational numbers. This is because the Cauchy convergence criterion only establishes the relationship between convergent sequences and Cauchy sequences. It is not suitable for proving that the set of rational numbers (Q) is incomplete and the set of real numbers (R) is complete.

But when we consider the perspective of limit closure, a connection can be established. If a sequence {Xn} converges to a value a, and if we restrict our discussion to the set of rational numbers (Q), then the Cauchy convergence criterion may not hold because {Xn} is a sequence of rational numbers, and a might be an irrational number.

------If this view is correct, then why do we insist on verifying the completeness of real numbers based on whether limits are closed, rather than demonstrating it through the closure of operations like square roots? After all, Pythagoras had already discovered irrational numbers when he tossed his students into the sea! Why wait until the development of mathematical analysis to prove that rational numbers are incomplete and real numbers are complete?


So, the completeness of real numbers is an important foundation for the development of calculus, but it's not the main goal of calculus research, is my understanding correct?

If there are any books that can help me understand the above issues, please recommend them in the comments. It's better not to involve too complex mathematics because I've found that many comments have already gone beyond my current level of learning. Please include ISBN numbers for the recommended books if possible.


2023-08-23 enter link description here Mikhail Katz provided a very helpful response.

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    $\begingroup$ Dedekind completeness of the reals (as an ordered space) is equivalent to completeness of the reals as a metric space yeah $\endgroup$
    – FShrike
    Commented Aug 19, 2023 at 8:10
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    $\begingroup$ True, (in)completeness in terms of least upper bounds is a separate concept from the (in)completeness in terms of Cauchy sequences. The former applies to partially ordered sets (posets), the latter to metric spaces. In case of the rationals, which are both a poset and a metric space, the process of completion yields the same thing ($\mathbb R$) and, being equivalent, it is just the matter of which one is taught first. $\endgroup$
    – user700480
    Commented Aug 19, 2023 at 8:13
  • $\begingroup$ As for the mathematical analysis (your last comment), it is not that much of an achievement to prove that $\mathbb Q$ is incomplete as to prove that $\mathbb R$ is complete. In other words, you can do calculus on it and you do not need to introduce infinitesimals (which was very important milestone to achieve as it made calculus rigorous). $\endgroup$
    – user700480
    Commented Aug 19, 2023 at 8:16
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    $\begingroup$ As for this sentence (presume from the book): "(the Cauchy criterion) ...is not suitable for proving that the set of rational numbers (Q) is incomplete and the set of real numbers (R) is complete.", I don't really know what "not suitable" means. That $\mathbb Q$ is incomplete follows very easily (say run the algorithm of halving the interval $[1,2]$ for approximating $\sqrt{2}$ and notice that the sequence you got is Cauchy). That $\mathbb R$ is complete is obvious if you define $\mathbb R$ as a Cauchy completion of $\mathbb Q$. $\endgroup$
    – user700480
    Commented Aug 19, 2023 at 8:26
  • $\begingroup$ @StinkingBishop Your point is: it wasn't the development of calculus that led humanity to discover the incompleteness of rational numbers. Humanity had already realized that rational numbers were "incomplete" long before, but they didn't know that real numbers were complete. Even in the time of Pythagoras, they only became aware of the incompleteness of rational numbers; nobody had rigorously proven the completeness of real numbers. The discovery of the square root of 2 highlighted the problem but didn't solve it. The square root of 2 alone couldn't prove the completeness of real numbers. $\endgroup$ Commented Aug 19, 2023 at 8:37

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