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• CommentRowNumber1.
• CommentAuthorUrs
• CommentTimeSep 28th 2012

I looked at real number and thought I could maybe try to improve the way the Idea section flows. Now it reads as follows:

A real number is something that may be approximated by rational numbers. Equipped with the operations of addition and multiplication induced from the rational numbers, real numbers form a number field, denoted $\mathbb{R}$. The underlying set is the completion of the ordered field $\mathbb{Q}$ of rational numbers: the result of adjoining to $\mathbb{Q}$ suprema for every bounded subset with respect to the natural ordering of rational numbers.

The set of real numbers also carries naturally the structure of a topological space and as such $\mathbb{R}$ is called the real line also known as the continuum. Equipped with both the topology and the field structure, $\mathbb{R}$ is a topological field and as such is the uniform completion of $\mathbb{Q}$ equipped with the absolute value metric.

Together with its cartesian products – the Cartesian spaces $\mathbb{R}^n$ for natural numbers $n \in \mathbb{N}$ – the real line $\mathbb{R}$ is a standard formalization of the idea of continuous space. The more general concept of (smooth) manifold is modeled on these Cartesian spaces. These, in turnm are standard models for the notion of space in particular in physics (see spacetime), or at least in classical physics. See at geometry of physics for more on this.

• CommentRowNumber2.
• CommentAuthorjin
• CommentTimeSep 11th 2020

Provide sketch of the content of the paper.

• CommentRowNumber3.
• CommentAuthorDavidRoberts
• CommentTimeSep 11th 2020

Moved the Pavlovic–Vaughan reference to the reference section, cited it, gave link to the published version, and put proper TeX syntax on the additions.

But I’m not sure of this sentence:

\ldots the result is simply the stream of $\mathbb{N}$: $\mathbb{N}\times \mathbb{N}\times \mathbb{N}\times \ldots$, which obviously can be identified with the real number set.

Really? Is it so obvious that this infinite product can be so identified?

• CommentRowNumber4.
• CommentAuthorMike Shulman
• CommentTimeSep 11th 2020

Looks more like Baire space to me.

• CommentRowNumber5.
• CommentAuthorDavidRoberts
• CommentTimeSep 12th 2020

I agree.

• CommentRowNumber6.
• CommentAuthorTodd_Trimble
• CommentTimeDec 21st 2020

The confusion with Baire space, noted in previous comments, has been rectified.

• CommentRowNumber7.
• CommentAuthorDavidRoberts
• CommentTimeDec 21st 2020

Changed the description as a locale a little. It was ambiguous in the previous version what the frame actually was, due to a plural/singular mismatch across two sentences.

• CommentRowNumber8.
• CommentAuthorOscar_Cunningham
• CommentTimeDec 22nd 2020
• (edited Dec 22nd 2020)

I changed the identification between $\omega\times\omega\times\dots$ and $\mathbb{R}^+$ to a nice one from Continued fractions and order-preserving homeomorphism that doesn’t require offsetting $\omega$ and $\mathbb{R}^+$.

I also corrected an error that said that $\alpha$ and $\beta$ were both monotonic. The function $\beta$ is not monotonic, but is monotonic when restricted to a domain where $\alpha$ is constant.

• CommentRowNumber9.
• CommentAuthorDavidRoberts
• CommentTimeDec 22nd 2020

Clarified that it is the underlying sets of Baire space and the real numbers that are isomorphic, since Schroeder-Bernstein is being invoked, and that is false in the category of topological spaces.

• CommentRowNumber10.
• CommentAuthorDavidRoberts
• CommentTimeDec 22nd 2020

Crossed-linked stream in the terminal coalgebra proof.

• CommentRowNumber11.
• CommentAuthorTodd_Trimble
• CommentTimeDec 22nd 2020
• (edited Dec 22nd 2020)

Oscar, you’re right; thanks.

• CommentRowNumber12.
• CommentAuthorTodd_Trimble
• CommentTimeDec 22nd 2020
• (edited Dec 22nd 2020)

However, I still like my original description, which works if we simply change from the product poset $\omega \times \mathbb{R}_+$ to the lexicographic one. We can keep yours, but I want mine reentered into the record.

Added: are you sure $\beta$ is monotonic? I get $\beta(1.9) \gt \beta(2.1)$. I think in fact it’s the very same issue as before.

1. I don’t understand what you mean about product vs lexicographic. If you go ahead and make the change I’ll look at what you wrote.

• CommentRowNumber14.
• CommentAuthorTodd_Trimble
• CommentTimeDec 22nd 2020

I’ll say it here first: if $x \leq y$, then $(\alpha(x), \beta(x)) \leq (\alpha(y), \beta(y))$ using the lexicographic order.

2. Right. I think the confusion was that you thought that my change to $\beta$ was to make it monotonic, whereas in fact I simply changed $\beta$ to allow the use of $\omega$ and $\mathbb{R}^+$ directly rather than having to replace them with $\mathbb{N}_{\geq 2}$ and $[1,\infty)$.

Then I unrelatedly changed the sentence that said that $\alpha$ and $\beta$ are monotonic to one which says that $(\alpha,\beta)$ is monotonic. And since we already mentioned in the definition of $F_1$ that the order on $\omega\times X$ is the ordinal product (which is lexicographic), I think the current wording already says exactly what you want.

• CommentRowNumber16.
• CommentAuthorTodd_Trimble
• CommentTimeDec 22nd 2020

3. You should remove my version then. I thought it was simpler, but it’s so similar to the other one that there’s no real gain in having both. If we want a second example there’s the function that sends a nonnegative real to the list of how many $1$s there are between each $0$ in its binary expansion.

• CommentRowNumber18.
• CommentAuthorTodd_Trimble
• CommentTimeJun 15th 2021

I tried to repair some wrong claims about terminal archimedean structures.

• CommentRowNumber19.
• CommentAuthorGuest
• CommentTimeMar 30th 2022
• CommentRowNumber20.
• CommentAuthorGuest
• CommentTimeApr 18th 2022

The pre-algebra real numbers:

The real numbers are an Archimedean ordered field such that for every real number $r$ in the interval $[0, 1]$, there is a sequence of integers $(a_n)_{n\in\mathbb{N}}$ bounded below by $0$ and bounded above by $9$ such that

$r = \lim_{n \to \infty} \sum_{i=0}^{n} \frac{a_i}{10^{i+1}}$

If there exists a natural number $N$ such that the sequence $(a_n)_{n\in\mathbb{N}}$ is periodic for $n \geq N$, then $r$ is a rational number.

• CommentRowNumber21.
• CommentAuthorGuest
• CommentTimeApr 18th 2022

That definition of the real numbers is incorrect: it is true for the integers and the rational numbers as well. One needs to additionally stipulate that for every sequence of integers bounded below by 0 and bounded above by 9, there is a real number $r$ such that the equation holds. Or one could simply include the structure of a set isomorphism between the unit interval and the type of sequences of integers bounded below by 0 and bounded above by 9.

• CommentRowNumber22.
• CommentAuthorHurkyl
• CommentTimeApr 18th 2022
• (edited Apr 18th 2022)

… with the appropriate corrections for trailing repeated 9’s vs repeated 0’s.

Incidentally, I may be misremembering, but I think the first rigorous proof I saw of the existence of the field of real numbers was actually given by defining the underlying set to be the decimal numbers (with its technicalities), and actually defined arithmetic via elementary school arithmetic, suitably modified to work for nonterminating decimals.

4. Gowers talks about that a bit here: https://www.dpmms.cam.ac.uk/~wtg10/decimals.html

• CommentRowNumber24.
• CommentAuthorGuest
• CommentTimeApr 18th 2022
Auke Booij proved that the set of infinite decimal representations is isomorphic to the set of Dedekind real numbers with a locator, and that the set of Dedekind real numbers with a locator is isomorphic to the set of Cauchy real numbers defined by sequences of rationals: https://arxiv.org/abs/1805.06781
• CommentRowNumber25.
• CommentAuthorGuest
• CommentTimeApr 18th 2022
Hurkyl, I think if one uses the epsilon-delta definition of a limit (using rational numbers or decimal numbers for epsilon and delta as common in predicative constructive mathematics), as well as the convergence of a geometric series, one could prove that infinite decimals with trailing repeated 0s are equal to other infinite decimals with repeated 9s, which I believe fundamentally only requires a proof that the infinite decimal consisting of all 9s is equal to 1.
• CommentRowNumber26.
• CommentAuthorGuest
• CommentTimeApr 18th 2022
Should note that the definition of the real numbers given above is an axiomatic definition of the real numbers, rather than a construction of the real numbers from infinite decimals, in the same way that one could define the real numbers as a Dedekind complete Archimedean ordered field, or a Cauchy complete Archimedean ordered field (in the sense of Cauchy nets), rather than the the Dedekind completion of the rational numbers or the Cauchy completion of the rational numbers.
• CommentRowNumber27.
• CommentAuthorHurkyl
• CommentTimeApr 19th 2022
• (edited Apr 19th 2022)

@25: My post #22 was a correction to the comment in #21 about a set isomorphism between the unit interval and decimal numerals.

• CommentRowNumber28.
• CommentAuthorGuest
• CommentTimeApr 19th 2022
Ah, well in which case the "appropriate corrections for trailing repeated 9’s vs repeated 0’s" is probably the limit equation in #21, with the limit of a sequence defined as its usual definition.
• CommentRowNumber29.
• CommentAuthorGuest
• CommentTimeMay 2nd 2022

@24: Auke Booij proved the result for signed digit infinite decimal representations. If every Cauchy real number has an unsigned infinite decimal representation in the sense you are talking about, then the limited principle of omniscience for natural numbers holds, as proven by Errett Bishop and other constructive mathematicians.

5. adding another definition of the real numbers

Anonymous