Questions tagged [ordered-fields]

Ordered fields are fields which have an additional structure, a linear order compatible with the field structure. This tag is for questions regarding ordered fields and their properties, as well proofs related to un-orderability of certain fields.

Filter by
Sorted by
Tagged with
2
votes
1answer
69 views

Is every ordered field a topological field?

Let $F$ be an ordered field and give $F$ an order topology. Then is $F$ a topological field (that is, are the operations \begin{equation} \begin{split} +&:F\times F\to F\\ -&:F\to F\\ \times&...
5
votes
1answer
134 views

A natural choice of 'maximal ordered subfield' of a field?

For any field $K$ of characteristic zero, its prime subfield $Q(K)$ has a natural ordering inherited from $\mathbb{Q}$. I am interested in finding out to what extent (if at all) this natural ordering ...
0
votes
1answer
59 views

Uniquely orderable subfields of $\mathbb{Q}_p$?

I have heard that, unlike $\mathbb{R}$, the field $\mathbb{Q}_p$ cannot be realised an ordered field. Is there any way to extend the natural ordering on $\mathbb{Q}$ to a larger subfield of $\mathbb{Q}...
6
votes
2answers
182 views

Are $\mathbb{R}$ and $\mathbb{Q}$ the only subfields of $\mathbb{C}$ with natural structure as ordered fields?

We know that $\mathbb{R}$ and $\mathbb{Q}$ have a unique structure as ordered fields with the usual order, and that $\mathbb{C}$ cannot be realised as an ordered field. Various non-trivial subfields ...
1
vote
1answer
41 views

If a field has unique ordering, must its subfields also have unique ordering?

Let $L$ be a field and suppose that there exists a unique total order $\leq$ on $L$ with respect to which $(L,\leq)$ is an ordered field. Now let $K$ be a subfield of $L$. Clearly $K$ is an ordered ...
1
vote
1answer
90 views

Showing that $i$ is neither negative nor positive [duplicate]

Show that $i$ is neither negative nor positive. Proof: Assume that $i<0$ $i×i>0×i$ Since $i<0$ $i^{2}>0$ Since $i^{2}:=-1$ So, $-1>0$ which doesn't hold and hence this is a ...
1
vote
0answers
23 views

Why must the field $K$ be ordered in the Inertia Theorem?

Page 11 of "Linear Algebra and Geometry" by Irving Kaplansky Theorem 8: Let $V$ be a finite dimensional inner product space over an ordered field $K$. Then $V$ can be written as an ...
4
votes
1answer
66 views

Real closure is the smallest real-closed field extension

Let $F$ be an ordered field, $R/F$ a real closure and $R'$ an arbitrary real-closed field extension of $F$. Need there be an (ordered) $F$-homomorphism $R\to R'$? Does this follow from the essential ...
2
votes
1answer
40 views

Cauchy completeness of an ordered field

Let $\Omega(t)=\Bbb{R}(t)$ denote all rational functions with real coefficients. Supposing $x(t),y(t) \in \Omega(t)$, a total strict order $<$ is defined where $x(t)<y(t) \iff \exists \, T \in \...
0
votes
1answer
41 views

Related to partially ordered field (poset) and total ordered field

(i) Let K be a field with the property that there is a positive natural number m such that $m * 1 = 0$. Show that there is no total order that gives K the structure of an ordered field. (ii) Show that ...
2
votes
1answer
99 views

$F(\sqrt x \mid x > 0)$ is formally real

In Jacobson's Basic Algebra IV, it is claimed in the proof of Theorem 8 page 285 that if $F$ is an ordered field then the extension $L = F(\sqrt x\mid x > 0)$ obtained by adjoining the square roots ...
4
votes
0answers
94 views

is this proof of 1>0 in an ordered field correct?

As I do not know what to write here, I write only that I tried to prove that $1>0$ using the ordered field axioms (Ordered Field Axioms). I simplified what the site said for shortness only. Try ...
1
vote
1answer
46 views

Is there a way to represent the relation $<$ on the real numbers without using multiplication?

I know for $x,y\in\mathbb R$ we have $x<y$ iff there exists a $z\in\mathbb R$, so that $y=x+z^2$. Is there a similar way to „prove“ $x<y$ using only $+$ and $=$? And if not, is there a more or ...
0
votes
2answers
117 views

What is a connex relation?.

I've been recently trying to teach myself some topology and in the book I'm reading there is the definition of an order relation which confuses me a lot . I've looked online for more information and ...
1
vote
0answers
21 views

Prove that a relation is a total order.

Consider the set $A=\{(x, y) \in \Bbb R\times\Bbb R \,|\, y \le 0\}.$ Prove that the following relation is of total order on $A.$ $$(x,y)\preceq(u,v)\iff (x<u) \vee [(x=u) \wedge (x^2+y^2 \ge u^2+v^...
0
votes
0answers
18 views

Ordered field order isomorphic

Let $K$ be an ordered field that satisfies the least upper bound property. Prove that there is a unique function $T:\mathbb{Q} \rightarrow \mathbb{Q}_K$ that is order isomorphic. That is, show that it ...
1
vote
3answers
53 views

Solutions in a field

Suppose $F$ is a field. Suppose further that $F$ is an ordered field. Consider the following equation $x^3 = b$, where $b \in F$. Prove that the equation above has, at most, one solution. My attempt: ...
1
vote
3answers
76 views

Why is the subspace topology same as the Order topology?

Hi i am reading Topology by Munkres and there in the proof of theorem 27.1 it is written that Given $a<b$, let $\mathcal A$ be a covering of $[a,b]$ by sets open in $[a,b]$ in the subspace ...
0
votes
1answer
34 views

Do there exists any Order Complete Field at all?

It is known that any two Order-Complete Fields are isomorphic. So there can exist at most 1 Order-Complete Field up to isomorphism. Is there a way to tell that there exists one Order Complete Field?
0
votes
1answer
25 views

Monotone Convergence Property $\iff$ Least Upper Bound Property

Let $\mathbb{F}$ be an arbitrary ordered field that satisfies Monotone Convergence Property then $\mathbb{F}$ is Order Complete. Order Completeness Property: If $S \subset \mathbb{F}$ is bounded ...
1
vote
0answers
16 views

Subfield of $\mathbb{R}$ that satisfies the Ratio Test must contain every reals.

I am reading James Propp's Real Analysis in reverse. There in pg. 12 he shows Ratio Test $\implies$ LUB Axiom. First he shows that Ratio Test $\implies$ Archimedean Property: By Consider the ...
0
votes
1answer
51 views

Every Archimedean Ordered Field is field isomorphic to a subfield of $\mathbb{R}$.

Let $\mathbb{F}$ be an archimedean ordered field. Then $\mathbb{F}$ is isomorphic to a subfield of $\mathbb{R}$. Let $x \in \mathbb{F}$, then define $S_x:=\{\frac{m}{n} \in \mathbb{Q}: \frac{m_{\...
1
vote
0answers
24 views

Proving order relation

Suppose $F = \mathbb{Q}^2$ is a field equipped with the following operations: Addition: Defined entrywise i.e. $$(x_1, x_2) + (y_1, y_2) = (x_1 + y_1, x_2 + y_2)$$ Multiplication: $$(x_1, x_2) \cdot (...
0
votes
1answer
25 views

Proving that a field cannot be an ordered field

Suppose $F = \mathbb{R}^2$ is a field equipped with the following operations: Addition: Defined entrywise i.e. $$(x_1, x_2) + (y_1, y_2) = (x_1 + y_1, x_2 + y_2)$$ Multiplication: $$(x_1, x_2) \cdot (...
2
votes
0answers
30 views

Topological Connectedness $\implies$ Order Completeness

Let $\mathbb{F}$ be an arbitrary ordered field. We define: $\textbf{Topological Connectedness:}$ If $\mathbb{F}=A \cup B$ where A,B are nonempty and open, then $A \cap B \neq \phi$. $\textbf{Order ...
0
votes
0answers
28 views

Convergence of Series in the Field of Formal Laurent Series.

This is a continuation of this question. I want to check if $(a_n) \to 0$, then $\sum\limits_{n=1}^\infty a_n$ is convergent. As @reuns stated in his answer to the question in his above link. If $(a_n)...
3
votes
1answer
90 views

$R((x))$ Set of formal Laurent Series is an ordered field which is Cauchy Complete.

Let $R$ be any ordered field. Define $R((x)):=\{\sum\limits_{k=-n}^\infty a_kx^k:a_k \in R \ , \ n \in\mathbb N\}$ Addition: $\sum_\limits{k=-n}a_kx^k+\sum_\limits{k=-m}b_kx^k:= \sum_\limits{k=-max\{n,...
2
votes
0answers
46 views

Is the scalar product defined always?

In a vector space, the scalar product has the property that $\langle v|v\rangle\ge 0$, where $0$ is the additive identity of that field. We use "greater than" and "equal" sign ...
0
votes
3answers
66 views

Does the set of all formal Laurent series form an ordered field?

Let $R$ be an ordered field. Consider the set $R[[x]]:=\{\sum_\limits{n \in \mathbb{Z}}a_nx^n:a_n \in R\}$ Addition: $\sum_\limits{n \in \mathbb{Z}}a_nx^n+\sum_\limits{n \in \mathbb{Z}}b_nx^n:= \sum_\...
0
votes
1answer
56 views

Monotone Convergence Property $\iff$ Order Completeness in an arbitrary ordered field.

I have found an answer here but having some doubts about the first answer(highest rep.) posted by Jonas Meyer. There in Claim 1. He states for all $\epsilon>0$, $U-\epsilon:=\{u- \epsilon: u \in U\}...
1
vote
2answers
34 views

Is Nested Interval Property $\iff$ Monotone Convergence Property in an ordered field?

Given any ordered field $R$ Can we conclude Monotone convergence Property $\iff$ Nested Interval Property? Monotone Convergence Property: Any monotone increasing sequence bounded above in $R$ is ...
3
votes
1answer
41 views

Is Nested Interval Property $\iff$ Order Completeness in any Ordered Field?

Let $R$ be an arbitrary Ordered Field. Order Completeness Property: If $S \subset R$ is bounded above, then $\exists \ c \in R$ that is an upperbound of S and for every upperbound $b$ of S, we have $c ...
2
votes
0answers
17 views

A corollary of the Nested Interval Property.

Let $R$ be an ordered field with the Nested Interval Property: If $I_1,I_2,...,I_n,...$ be a collection of nested closed intervals, i.e $I_1 \supseteq I_2 \supseteq ... \supseteq I_n \supseteq ...$ ...
0
votes
1answer
59 views

How can you prove $\mathbb Q$ is Archimedean directly from its axiomatization as the minimal ordered field?

The Archimedean property for $\mathbb R$ follows from the axioms of complete ordered field. $\mathbb Q$ is often characterized by stating it to be the minimal ordered field (with $0\ne 1$). But this (...
1
vote
2answers
59 views

Relaxation of valuation axioms?

Let $K$ be a field, and $v:K\rightarrow \mathbb{R}$ be a valuation, i.e. a function such that, for all $x,y\in K$: $v(x)\geq 0$, $v(x+y)\leq v(x)+v(y)$ (if $v$ is Archimedian, alternatively $v(x+y)\...
0
votes
1answer
18 views

Formal proof about ordered fields

Let $(\mathbb F,\mathbb F^{+})$ an ordered field. Are the following statements true? Prove (or disprove). (i) For every element $x \in \mathbb F$ there are two other elements in $\mathbb F$ which are ...
2
votes
1answer
22 views

Every cofinal real closed subfield is dense?

Consider a real closed subfield $F_0$ of a real closed field $F_1$ that is cofinal. Here, "$F_0$ is cofinal in $F_1$" means "For each $x \in F_1$, there exists $y \in F_1$ such that $x &...
0
votes
1answer
32 views

Assume $x,y \in F$ and F is an ordered field, if $x<0$ and $y<0$, then $xy>0$.

I was thinking since $x<0$ and $y<0$, $-x>0$ and -y<0 by adding the additive inverse. Then we multiply -x and -y to get $(-x)(-y)>0$ since $-x>0$ and $-y>0$, but I feel like this ...
3
votes
1answer
136 views

What is Arithmetic Continuum

I recently tried understanding Surreal Numbers on a more meaningful level. Along the way I found this answer to a related question. The accepted answer and the paper it suggests contain the term "...
0
votes
1answer
30 views

Ordered field with two embeddings

Do there exist any two ordered fields $F_1,F_2$ such that $F_1$ admits two distinct ordered field embeddings $f,g:F_1\rightrightarrows F_2$? Note that $\mathbb{Q}[\sqrt 2]\rightrightarrows\mathbb{R}$ ...
0
votes
1answer
57 views

$\mathbb{R}(x)$ as non o-minimal structure

Let us consider the field of real rational functions in one variable $\mathbb{R}(x)$ as an ordered field with $ x > r$ for all $r \in \mathbb{R}$. have to show that $\mathbb{R}(x)$ is not o-...
0
votes
2answers
55 views

Spivak's Calculus Chapter 28-1 (Fields)

Question in Textbook The question defines a set F and then defines the operations on F through a table. In this chapter, Spivak outlines the properties that operators on a field must have. In the ...
0
votes
1answer
46 views

Showing that it is impossible to prove a statement from some facts.

I am loving the question that is Show that the Completeness Axiom can not be proved using the 10 axioms of an ordered field. The 10 axioms are 5 axioms for ordering properties and, commutativity, ...
3
votes
3answers
124 views

Ordered field and all Cauchy sequence eventually constant

What will be an example of an ordered field in which every Cauchy sequence is eventually constant? I think an example exists. See math.uga.edu/~pete/Clark-Diepeveen_PLUS.pdf. This proves that all ...
1
vote
0answers
19 views

Ordered Field with BW Property but not having LUB property

In an Ordered Field, LUB Property $\implies$ BW Property (i.e Every bounded sequence has a Convergent Subsequence.) Is the converse true?
3
votes
0answers
97 views

Isomorphism of hyperreal fields viewed as extensions of the real field

Crossposted on MathOverflow: https://mathoverflow.net/q/368381/461 Let $A$ be the $\mathbb R$-algebra of all $\mathbb R$-valued functions on $\mathbb N$, that is $$ A=\mathbb R^{\mathbb N}=\prod_{n\in ...
0
votes
0answers
15 views

Solving conjugation equations in bi-ordered groups

A bi-ordered group is a group $G$ equipped with a partial order $<$ with: $f<g\Longleftrightarrow h f i < h g i$ for all $f,g,h,i \in G$. In particular this is the case for groups of strictly ...
0
votes
1answer
55 views

How to prove: Any positive real number power of positive real number is positive.

I have a question regarding how to prove: Any positive real number power of a positive real number is positive. I have shown that any positive rational power of positive real numbers is positive. I ...
3
votes
1answer
96 views

Is there an ordered field with distinct subfields isomorphic to the reals?

Is there an ordered field with distinct subfields isomorphic to the field $\mathbb R$ of real numbers?
1
vote
0answers
31 views

Fields non-trivially seen to be isomorphic to the real numbers by the uniqueness of a complete ordered field

It is a well-known fact from introductory real analysis that $\mathbb{R}$ is the unique Dedekind-complete ordered field, up to isomorphism. Hence, we are in a sense justified in an abstract definition ...

1
2 3 4 5
7