14
votes
1answer
367 views
Why do rigid spaces have “not enough points”?
In Brian Conrad's notes
here for the 2007 Arizona winter school, bottom of p18, he says that there is an affinoid rigid-analytic space and a sheaf of abelian groups on it equipped …
1
vote
0answers
110 views
Asymptotics vs Puiseux series
Define asymptotic as a class of sequences {$ x_i$},$_{i\in\mathbb N}$ modulo equivalence {$x_i$}={$y_i$} if $\lim_{i\to\infty} (x_i/y_i)=c\in\mathbb R,c\ne 0$.
More, we define $X= …
8
votes
1answer
280 views
bornological vector spaces over a non-archimedean field
Let $k$ be a complete non-archimedean field. In definitions I have seen of bornological vector spaces over $k$ there are usually some extra assumptions on the non-archimedean field …
2
votes
2answers
315 views
Non-archimedean group over the reals
Hi there,
I have a totally ordered group $(\mathbb{R};\leq,\oplus,0,-)$ with the reals as base set satisfying monotonicity, i.e.
for all $x,y,z$ we have that if $y\leq z$ then $x\o …
8
votes
2answers
464 views
Does Rolle’s Theorem imply Dedekind completeness?
I think the answer to the title question is "yes", but Gerald Edgar, in his comment on http://mathoverflow.net/questions/87640/does-antidifferentiability-of-continuous-functions-im …
0
votes
1answer
186 views
Non-Archimedean non-standard models for R
Let $\langle \mathbb{R}, 0, 1, +, \cdot, <\rangle$ be the standard model for $R$, and let $S$ be a countable model of $R$ (satisfying all true first-order statements in $R$). Is …
8
votes
0answers
174 views
Does antidifferentiability of continuous functions imply Dedekind completeness?
Let $R$ be an ordered field, and let $I$ be {$x \in R: a < x < b$} for some $a < b$ in $R$. Define notions of $R$-continuity and $R$-differentiability for functions $f : …

