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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.
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?
Looks more like Baire space to me.
I agree.
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.
Oscar, you’re right; thanks.
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.
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.
I’ll say it here first: if $x \leq y$, then $(\alpha(x), \beta(x)) \leq (\alpha(y), \beta(y))$ using the lexicographic order.
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.
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.
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.
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.
… 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.
Gowers talks about that a bit here: https://www.dpmms.cam.ac.uk/~wtg10/decimals.html
@25: My post #22 was a correction to the comment in #21 about a set isomorphism between the unit interval and decimal numerals.
@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.
added pointer to:
working on polishing-up the list of references
among other things, i have added publication data to this item:
added pointer to:
Added reference
Anonymouse
Clarify which weak countable choice.
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