Let $(k,+,.,0,1,<)$ be an ordered field. In the folowing definitions, all numbers and notions are derived from the ordered field structure of $k$, and $a < c$ are generic elements of $k$. Define the limit of a function $f: ]a;b[ \cup ]b;c[ \longrightarrow k$ at $b$ as the element $l$ of $k$ if there is […]

It is well known that every continuous injective map $\mathbb{R}\rightarrow\mathbb{R}$ is monotone. This statement is false for maps $\mathbb{Q}\rightarrow\mathbb{Q}$. (That is becaus $\mathbb{Q}$ is not complete. You can change from increasing to decreasing and vice versa in each irrational “hole”). Is it true that every homeomorphism of $\mathbb{Q}$ is monotone?

Is it possible to find totally ordered fields $K$ and $L$, and a map $f: K \to L$ that is an isomorphism of the ordered additive group structures such that $f(1)=1$, but which is not an isomorphism of fields? I know that this is impossible if $K$ or $L$ is a subfield of the real […]

Is there any ordered field smaller that the set of real numbers in which we can do calculus, also with many restrictions ? If not why ?

Given an ordered field $F$, the following two statements are equivalent: $F$ has the Least-Upper-Bound property. $F$ is Archimedean and $F$ is “sequentially complete”/”Cauchy complete” (all Cauchy sequences in $F$ converge). For some reason I have been unable to find a proof of this result on the web.

I came across this question in a calculus book. Is it possible to prove that an ordered field must be infinite? Also – does this mean that there is only one such field? Thanks

Question: Explain the construction below (taken directly from Counter Examples in Analysis): An ordered field is a field $F$ that contains a subset $P$ such that $P$ is closed with respect to addition and multiplication and exactly one of the following are true: $$x \in P; x = 0; -x \in P$$ The set $F$ […]

I want to show that if an ordered field $X$ has the least upper bound property (meaning, every nonempty set $E$ which is bounded above has $\sup E \in X$), then it is complete (meaning, every Cauchy sequence converges in $X$). I know the converse is not true, but how do I prove this direction?

I’m going through the first chapter in a text on real analysis, which contains preliminaries on ordered fields, the real numbers, etc. Supposedly I had learned about such things already, in calculus, but I thought it wouldn’t hurt to go over it again. Up until the paragraph which is the subject of my question, everything […]

I am reading baby Rudin and it says all ordered fields with supremum property are isomorphic to $\mathbb R$. Since all ordered finite fields would have supremum property that must mean none exist. Could someone please show me a proof of this? Thank you very much, Regards.

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