@TobiasKildetoft The problem is that, I don't think passing through a nonsingular curve would lead to anything meaningful. It's endowed with subspace topology, so it doesn't matter if you consider the perturbation a path which intersects with the subspace of nonsingular curves, say.
@TobiasKildetoft My guess is that, the set with at least two singular points, or a degenerate singular point, is the Zariski closure of the set with at least two singular points.
Suppose $L_d$ is the linear system of curves of degree $d$. I think we can start with the canonical projection $((\mathbb P^1\times\mathbb P^1)\setminus\Delta)\times L_d\to L_d$, and restrict to a subvariety of $(p,q,C)$ where $C$ is singular at $p$ and $q$.
@Axoren Yes. For example, Bourbaki starts with topological vectors spaces over a complete valued division ring.
@Danu Yeah. I was hoping to learn the varieties of ways to construct space-filling curves and if there was some way to optimize the computation of them. I'm finding a lot of efficient ways to compute very specific curves that don't suit my needs, and everything else I find is proof-of-existence.
the motivation behind sheaves is to axiomatize what sort of "functions" on a topological space behave in a nice enough way. And then realizing that these need not really be functions at all
> Despite its similarity to "étalé", the word étale [etal] has a different meaning both in French and in mathematics. In particular, it is possible to turn E into a scheme and Ï€ into a morphism of schemes in such a way that Ï€ retains the same universal property, but Ï€ is not in general an étale morphism because it is not quasi-finite. It is, however, formally étale.
@DanielFischer we say that $(f_n)$ has a Cauchy subsequence say $(f_{\varphi(n)})$ iff $\forall \varepsilon>0, \exists \varphi(n)_0>0, \forall p,q>\varphi(n)_0, d(f_{\varphi(p)},f_{\varphi(q)})<\varepsilon$?
@Vrouvrou No, having a Cauchy subsequence is a stronger condition. But if this weaker condition is not satisfied, then the sequence a fortiori has no Cauchy subsequence.
@TobiasKildetoft I get it. So if I want to find the order of $g^i$ for say $2 \le i \le 13$ I do that for each of them or is there a more systematic approach I could use?
@Vrouvrou If $(f_{n_k})$ were a Cauchy subsequence of $(f_n)$, what would that imply about $d(f_{n_k}, f_{n_m})$? Is what it would imply compatible with the fact that $d(f_n,f_m) = \sqrt{\pi}$ for all $n \neq m$?
so to prove that $(f_n)$ has no adherent values we prove that $(f_n)$ has no a Cauchy subsequence , we suppose that $(f_n)$ has a Cauchy subsequence $(f_{n_k})$ it means that $\forall \varepsilon>0, \exits k_0>0, \forall p,q>k_0, d(f_{n_p},f_{n_q})<\varepsilon$ but we have that $d(f_{n_p},f_{n_q})=\sqrt{\pi}$ then contradiction so $(f_n)$ has no adherent value
We are over $\mathbb{F}_2$ and we have $x,y$ the number of non-zero componenents of which is a multiple of 4. Always if we add x to y, the result will have a multiple of 4 non-zero componenets, right? @DanielFischer
It's from coding theory... We have self-dual binary [n,k,d] code and we want to show that $\{ \{ c \in C: ||c|| \equiv 0 \mod 4\}$ is a linear subspace of $C$. @DanielFischer
The length of all the words of the code is n, the words that span the code are contained at the generator matrix, the dimension of which is k and d is the minimum weight of a word (weight: number of non-zero components of a word). @DanielFischer
Geia @NikolaosSkout
@DanielFischer A dual code $C^{\perp}$ of a code $C$ contains the words x for which $\langle x,y \rangle=0$ for all $y \in C$. So for a self-dual code it holds that $\langle x, x \rangle=0$ for all $x \in C$.
@Danu: Maybe it's worth noting the following. While the "constant presheaf" makes perfect sense (just $F(U) = A$ for all $U$), it's not a sheaf: if there is a disconnection $A \cup B$ of the space $X$, the gluing axiom says that $F(X) = F(A) \oplus F(B)$, so the restriction map $F(X) \to F(A)$ is not an isomorphism.
It's not a sheaf on a disconnected space, that is.
On a connected space, "locally constant functions" are just constant functions, and the constant presheaf is a sheaf.
OK, so how do you sheafify a presheaf? You pass to the etale space (in this case, the space $X \times A$, where $A$ has the discrete topology) and take sections of this bundle. Then $F(U)$ isn't the set of constant functions to $A$ - it's locally constant functions.
It's just, even if you only care about more obvious stuff, like "How many vector bundles are there?" or "What's the cohomology of my space?", the way you talk about this is sheaf cohomology, and makes some things much easier to say.
So very soon you'll understand why $H^1(Y;\Bbb R^\times)$ (and here the sheaf is continuous functions to $\Bbb R^\times$ - not locally constant or anything like that) classifies real line bundles. There's an exact sequence of sheaves $0 \to \Bbb Z/2 \to \Bbb R^\times \to \Bbb R \to 0$ (the third map is exponentiation)
This gives you an exact sheaf cohomology sequence $H^0(X;\Bbb R) \to H^1(X;\Bbb Z/2) \to H^1(X;\Bbb R^\times) \to H^1(X;\Bbb R)$. $\Bbb R$ is a "fine sheaf", henece $H^(X;\Bbb R) = 0$, and it's also easy to see that the first map is zero. Hence $H^1(X;\Bbb Z/2) \to H^1(X;\Bbb R^\times)$ is an isomorphism, and real line bundles are classified by $H^1(X;\Bbb Z/2)$.
For $X = S^1$ that means there's two of 'em, and you can write down two different ones: the trivial bundle and the Mobius bundle. So that's all the line bundles on the circle.
So, you could probably write all of this down without ever saying the word sheaf or sheaf cohomology. But it gives you a language to make such arguments systematic.
I have a boolean algebra expression that I have to simplify. I also have the answer from a logical calculator. However I can't seem to get from the problem to the simplified answer using only standard boolean rules (I can't use sum-of-product reduction). Anyone interesting in helping?
