28
votes
Cauchy reals and Dedekind reals satisfy "the same mathematical theorems"
Here is a way to do this in ZFC. Similar ideas work in a bunch of other contexts.
First, given any set $A$ in the universe of sets $V$ we can form the set theoretic universe $V(A)$ by mimicking the ...
- 43.7k
21
votes
Accepted
Cauchy reals and Dedekind reals satisfy "the same mathematical theorems"
Here's a low-tech way to look at it, which to me seems perfectly convincing.
Let C be some implementation of the reals via Cauchy sequences and D be some implementation of the reals via Dedekind cuts....
- 504
18
votes
Cauchy reals and Dedekind reals satisfy "the same mathematical theorems"
From a logical viewpoint, this has nothing to do with platonism, ZFC, or the cumulative hierarchy.
$
\def\nn{\mathbb{N}}
\def\zz{\mathbb{Z}}
\def\qq{\mathbb{Q}}
\def\rr{\mathbb{R}}
\def\cc{\mathbb{C}}
...
- 2,733
15
votes
Accepted
Is this theory the complete theory of the real ordered field?
It is not. Using set forcing, we can add 'undefinable' reals in a controlled manner, while keeping complexity of parameter-free definable sets low.
Specifically, let $M$ be a countable $ω$-model of ...
- 6,112
14
votes
Are archimedean subextensions of ordered fields dense?
Let $E$ be the real closure of $\mathbb{Q}(x, y) = (\mathbb{Q}(x))(y)$, with order given by $x > \mathbb{Q}$ and$y > \mathbb{Q}(x)$. In other words, positivity on $\mathbb{Q}(x, y)$ is ...
- 4,961
13
votes
Accepted
Decidability of a first-order theory of hyperreals
Yes, the theory is decidable.
If $F$ is an ordered field and $R\subseteq F$ a non-cofinal subfield, then
$$O=\{x\in F:\exists u\in R\:(-u\le x\le u)\}$$
is a convex valuation ring of $F$, with maximal ...
- 44.8k
13
votes
Accepted
Is "All hyperreal fields $C(\mathbb{R})/M$ are isomorphic" independent of ZFC+$\lnot$CH?
In On ultra powers of Boolean algebras (Topology Proceedings 9 (1984) 269-291) Alan Dow proved (Corollary 2.3) that $\neg\mathsf{CH}$ implies there are two fields of the form $C(\mathbb{N})/M$ that ...
- 9,910
12
votes
Are radicals dense in the real closure of an ordered field?
The answer is negative, and your example of $F$ not being dense in $R$ gives an example of this as well. Let $\alpha=\sqrt{\omega}+1,\beta=\sqrt{\omega}+2$. Then $\alpha^n>\omega^{n/2}+n\omega^{(n-...
- 27.4k
12
votes
Cauchy reals and Dedekind reals satisfy "the same mathematical theorems"
I see a lot of confusion in the comments about what you are asking and why. The way I interpret your question is this:
Given the fact that mathematical statements about (or involving) the reals ...
- 2,490
12
votes
Cauchy reals and Dedekind reals satisfy "the same mathematical theorems"
There is no need to even go as far as $\mathbb{R}$ for an example of this type of phenomena. Even $\mathbb{Z}$ could be defined as different sets in ZFC. Let $\omega$ be the first infinite cardinal, ...
- 18.2k
11
votes
Accepted
Are archimedean subextensions of ordered fields dense?
Let $F$ be any real-closed field of uncountable cofinality. That is, every countable subset of $F$ is bounded. One can make such a field in a process of $\omega_1$-many field extensions; alternatively,...
- 225k
9
votes
Accepted
Is there a complete characterization of ordered fields without definable proper subfields?
This is an interesting question. We know some things about this, but we do not have a characterization of fields with this property. As Wojowu says above the restriction to countable fields doesn't ...
- 1,639
9
votes
Is this theory the complete theory of the real ordered field?
Noah Schweber suggested on math stack exchange that $\mathbb Q^{alg}(r)$ could satisfy this weak least upper bound property for some fixed transcendental number $r$, say $r=\pi$. In this case, it ...
- 137k
8
votes
Accepted
Which ordinals can be embedded into an ordered field?
This parameter of a field does not equal any its common cardinal characteristic that I could think of, though it is related in several ways.
Let me first introduce some notation. Assume $F$ is an ...
- 44.8k
8
votes
Which ordinals can be embedded into an ordered field?
Let me address merely the suggestion the OP makes in the comment: whether this ordinal can be specified from the cofinality of the field.
The answer is no, because any ordered field $F$ can be ...
- 225k
7
votes
Accepted
On a completeness property of hyperreals
This is also called Cauchy-completeness, and it coincides for non-Archimedean ordered fields with the natural valuation to the valuation-theoretic notion of completeness. Also, this is the same as ...
- 2,367
6
votes
Cauchy reals and Dedekind reals satisfy "the same mathematical theorems"
I think it is essentially the Frege–Hilbert controversy on the nature of mathematical axioms, cf. e.g. https://plato.stanford.edu/entries/frege-hilbert/ :
The central difference between Frege and ...
- 167
6
votes
Accepted
How do fractional tensor products work?
I'm not going to try to match Tao's notation.
I will use the word line to mean a one-dimensional real vector space.
Suppose that $L$ is a line. Consider the line $L^{\otimes 2}$. It has the ...
- 53.1k
6
votes
How do fractional tensor products work?
You can define the square root of a vector space $V$ simply as a vector space $U$ with a distinguished linear isomorphism $L: U\otimes U \rightarrow V$.
You then want the following universal property ...
- 27k
6
votes
Formally real fields with unique non-Archimedean ordering
Yes, such fields exist. Let $T$ be the first-order theory in the language of fields with an extra constant $c$, axiomatized by
the axioms of fields,
every element or its negative is a sum of 4 ...
- 44.8k
6
votes
Accepted
Is there an exponential map on (Hahn) ordered fields?
There is no such exponential map. This was demonstrated in:
F.-V. Kuhlmann, S. Kuhlmann, S. Shelah, Exponentiation in power series fields, Proc. Amer. Math. Soc. 125 (1997) 3177–3183.
- 5,915
6
votes
Accepted
Archimedean ordered fields without maxima and minima in constructive mathematics
In the topos of sheaves over $\mathbb{R}$ we can construct a sub-ordered field of the Dedekind reals that does not have meets and joins. We recall that we can explicitly describe the Dedekind reals in ...
- 3,846
5
votes
Accepted
Completing class-sized Fields
I suspect the following should address the above stated queries.
Let $A$ be an ordered field whose universe is a proper class of NBG (which I take to include Global Choice).
(i) A real-closure of $A$...
- 5,915
5
votes
Are archimedean subextensions of ordered fields dense?
This is similar to user44191's answer but I want to put the emphasis on the fact that there is no reason that the cofinality of $F$ in $E$ (which is what ou call [$E$ is $F$-archimedean]) imply the ...
- 2,367
5
votes
Is this theory the complete theory of the real ordered field?
Edit: As Emil Jeřábek has pointed out in the comments, there is an error in the paper linked below, and the example does not work. I'll leave the answer here to document the error and add to the ...
- 4,431
5
votes
Proper definition of ordered field in constructive mathematics
The word "constructive" has many variations. Some might even say that it's a moving target. This is not a one-size-fits-all situation, so I would contend that maybe the proper definition of ...
- 43.7k
4
votes
Is this theory the complete theory of the real ordered field?
In my hasty first reading I did not notice the stipulation about parameter-freeness; I am leaving this answer "only for the record".
The answer to the question is in the positive, provided &...
- 17.1k
4
votes
Is this theory the complete theory of the real ordered field?
This is a non-answer too long for a comment.
If $K$ is an ordered field that satisfies the least upper bound property for sets definable without parameters, then suprema of sets definable with ...
- 44.8k
3
votes
Accepted
'Smallest' subfield of the Surreals which is isomorphic to the Surreals as an ordered group
$\DeclareMathOperator{\Noo}{\mathbf{No}}$This might actually be a dead end.
This is because if $F$ is isomorphic as an ordered group to $\Noo$, then their value classes under natural ordered group ...
- 2,367
Only top scored, non community-wiki answers of a minimum length are eligible