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awarded  Yearling
Apr
21
answered $p$-adics with the least upper-bound property
Apr
21
comment $p$-adics with the least upper-bound property
No worries. I'm glad at least someone cares! If by an infinite $p$-adic you mean something like $-1=(p-1)+(p-1)p+(p-1)p^2+\cdots$, then, at least in this case, $\sup \{ 0,-1\} =0$. In particular, this ordering has the odd property that, while the sum of any finitely many positive numbers is positive, the sum of infinitely many positive numbers can be negative. Or the same statement without my language: the sum of infinitely many absolute values can be the additive inverse of an absolute value.
Apr
20
comment $p$-adics with the least upper-bound property
No. They are not comparable with 0. In particular, this is not a total order.
Apr
20
comment $p$-adics with the least upper-bound property
@John Brevik What do you mean? If $-x$ and $-y$ are both finite sums of powers of $p$, then $-(x+y)$ is likewise a finite sum of powers of $p$, no?
Apr
20
asked $p$-adics with the least upper-bound property
Apr
19
revised Non-closed ideals in $C^*$-algebras
edited body
Feb
26
awarded  Popular Question
Feb
21
accepted Non-closed ideals in $C^*$-algebras
Feb
21
revised Non-closed ideals in $C^*$-algebras
edited body
Feb
21
asked Non-closed ideals in $C^*$-algebras
Feb
19
accepted Exact sequence of groups to exact sequence sheaves
Feb
18
awarded  Tumbleweed
Feb
14
awarded  Promoter
Feb
12
answered Distribution theory book
Feb
11
asked Exact sequence of groups to exact sequence sheaves
Jan
18
awarded  Nice Question
Dec
16
asked Cartier divisors of schemes
Dec
3
comment Using the Fourier integral theorem to evaluate the improper integrals
Consider $\frac{\mathrm{d}}{\mathrm{d}p}\left[ \frac{1}{\sqrt{2\pi}}\int \mathrm{d}x\, \frac{\mathrm{e}^{-\mathrm{i}xp}}{x^2+1}\right]$. Is this enough of a hint?
Dec
1
revised prove that $HOM_R(\bigoplus _{s\in S}R_s ,M)$ and $\prod_{s\in S}M_s$ are isomorphic as $R$ modules.
added 18 characters in body