@Liad But in my case it's different since I'm doing a Grande École. I had to go through two years of CPGE (Classes Préparatoires aux Grandes Écoles) or prepa, then take a competitive exam, and get in a school
Yeah I kinda forgot I'm adding inverses for the stuff outside $\mathfrak p$ because the multiplicative set isn't $\mathfrak p$ but its complement @Mathei :P
Sanity check: since stalks are local if $U\subseteq X$ is open then $\mathcal O_{X,x}=\mathcal O_{U,x}$ ($\mathcal O_U$ is just the restriction of the structure sheaf $\mathcal O_X$ to $U$)
Because the stalk at $x$ is made of equivalence classes of functions agreeing in a nbhd of $x$ so restrincting to an open set (containing $x$ of course) makes no difference
I should go to sleep and stop saying wrong things :P So in general if I have a directed limit of $A_i$ where $i$ ranges over some direct set $I$ and I take the direct limit of $A_j$ where $j$ ranges over some cofinal subset of $I$ I should get the same thing
There are two inequivalent definitions around: a directed set is a preorder in which every two elements have an upper bound or a directed set is a poset in which every two elements have an upper bound
According to wikipedia the former is enough to do direct limits
Hmm actually it's not really clear which definition of directed set are they using on the wiki page for direct limit. I totally support defining a directed set as a poset with extra structure though
even with the first definition, it's basically equivalent to the other one for the purpose of limits: let $(X,\leq)$ be a preordered set, then choose a system of representatives for the equivalence relation $a \sim b :\Leftrightarrow a \leq b \land b \leq a$, then that subset with the restricted preorder is a partial order and is cofinal, so the limits are isomorphic
@LeakyNun How do you like the proof for the claim "if $A \neq \varnothing$, then $\exists a \in A$", that goes like this? consider $A$ as a family indexed over one element, then there exists an element in $A$ by the axiom of choice
@Astyx this is my attempt 1) L(f)=Sup{L(P;f) : P in P(I)} this is definition We want L(f) = Sup{L(P,f) : P in P*} P* is subset of P(I) Hence {L(P;f) : P in P*} is subset of {L(P;f) : P in P(I)} hence Sup{L(P;f) : P in P*} <= Sup{L(P,f) : P in P(I)}
Now for all P in P(I) we can add 0 to P and obtain P' in P* such that P' is a refinament of P, hence L(P;f) <= L(P',f) and by definition L(P',f) <=Sup {L(P*,f) : 0 in P*}:=L*(f)
So f(x,y)= ( sqrt ( x^2+y^2 ) - x ) / (y) is homogeneous not either ? That's why I'm asking all of these cuz I've been told that the function above is in fact homogeneous.
If you want, I can send you what I plan to present. I don't go too deep into it because it's a sciences talk, and I don't go too fine with the method (I think I discretized to something like 12 x 20?), but the essence of it is there.
@Daniel: Are you still working on upper/lower sums if you refine the partition, or on something else? I can't figure out what you're talking about by scrolling back.
But if you want to know, He's trying to prove that the lower sum can be equivalently defined as the sup of $\left\{L(p,f) : p \text{ is symmetric}\right\}$
