# Tag Info

6

You are right, the zero scheme of a section is not the complement of the support. In other words, the condition is not "$s_x=0$ in $\mathcal F_x$", but rather "$s(x)=0$". To answer your question, I have to make sense of the latter expression. Locally around $x$, a section $s\in \Gamma(X,\mathcal F)$ is represented by an $r$-tuple of regular functions ...

5

The function $f(x)=a^x$ for $a\geq 0$ is uniquely determined by these three properties: $\forall x,y \in\Bbb R\left[f(x+y)=f(x)f(y)\right]$ $f$ is continuous at at least one point $f(1)=a$ What Martín-Blas is trying to convey is that, once you agree that the fundamental property that an 'exponential-like' operation $E:\Bbb R\rightarrow \Bbb R$ has is ...

4

First notice that if $E$ is a linear space then its dual $E^*$ is the linear space of the linear forms $$\varphi: E\rightarrow \Bbb K$$ Now if $\Psi\in \operatorname{hom}(V,W)$ i.e. a linear transformation $\Psi:V\rightarrow W$ then $\Phi(\Psi)\in \operatorname{hom}(W^*,V^*)$ since $$\underbrace{\Phi(\Psi)(\varphi)}_{\in ... 4 IMO, the only hard part about this problem is wrapping your head around the notion of a function-valued function. Or worse, this problem is about a function-valued-function-valued function of functions. When I first started learning about these things, I managed by writing with a very precise syntax, so I could see exactly what type everything is, and ... 4 From what I understand, your question comes down to why a set being ordered does not imply it being countable. The slightly subtle notion here is the difference between cardinal and ordinal numbers. In general, there are two standard ways to compare the sizes of two sets. You can construct a bijection (as you've probably seen), or you can construct an ... 4 Let V be the total space of \mathcal{F}, i.e. the global spectrum of the quasicoherent sheaf of algebras \text{Sym}(\mathcal{F}^{\vee}). There is a natural projection V \to X. Then a global section of \mathcal{F} can be thought of as a morphism s : X \to V such that the composition X \to V \to X is the identity on X (literally a section of ... 4 Yes, we could also define the strong resp. weak operator topologies as the initial topology with respect to the evaluation maps p_x, where Y is endowed with the strong resp. weak topology. For the weak operator topology that is obvious from the transitivity of initial topologies, since the weak topology is just the initial topology with respect to the ... 3 This is in general not possible. First of all, whatever solution you get will not be unique, as you indicate, in the sense that global translations and rotations of the set of points will also be a solution to the problem. Beyond that, there are two possible pitfalls: The set of "distances" can be inconsistent if set wrong. At the very least it must obey ... 3 There is no difference. It is easy to prove that the statement For each \epsilon>0, there exists N\in\mathbb N so that for each n>N, |x_n-a|<\epsilon is equivalent to the statement For each \epsilon>0, there exists N\in\mathbb N so that for each n>N, |x_n-a|\leq\epsilon. 3 Abstractly, an "ordinal number" is an order type of well-orderings -- that is, intuitively whichever "thing" (order-)isomorphic well-ordered collections have in common is their ordinal number. The Von Neumann ordinals are a popular way of choosing one concrete set-theoretical object to represent each (set-sized) ordinal number. This identification is so ... 2 It's clear that |x_n-a|<\varepsilon implies |x_n-a|\leqslant\varepsilon. To see it conversely, note the condition "for all \varepsilon>0". Thus, since the statement "for all \varepsilon>0 ... |x_n-a|\leqslant\varepsilon" holds for arbitrary \varepsilon, it holds also when we replace \varepsilon by \varepsilon/2 throughout. That is, ... 2 I believe you are talking about a hereditary property. From Wikipedia: “In mathematics, a hereditary property is a property of an object, that inherits to all its subobjects, where the term subobject depends on the context.” 2 Context is important in mathematics. In some context, "ordinal number" represents an equivalence class of a well-ordered set when considering the equivalence relation to be order isomorphism. Sometimes, the context demands this to be a set, so we find a way to "trim the equivalence class" into a set, or even pick one canonical representative -- as in the ... 2 I disagree that these definitions are different - they are just the same definition written in three different ways. The first definition is what you will see in most textbooks; I would consider it to be the "standard" definition. To parse the second definition, recall that a sequence v_k of vectors in Euclidean space converges to v if the sequence ... 2 Many concepts in mathematics can be captured by the idea of a "set with structure" - there is some underlying set S of objects we're interested in, and then we endow or equip S with a "structure" (a vague word including, e.g., operations, topologies, orderings, etc.) by doing this: We form the ordered pair (S,\,\mathsf{structure}). By an abuse of ... 2 YES it is equivalent. If \langle x,y\rangle=\langle Tx,Ty\rangle, for x=y:$$\|x\|^2=\langle x,x\rangle=\langle Tx,Tx\rangle=\|Tx\|^2,$$and hence \|x\|=\|Tx\|, for all x. If \|x\|=\|Tx\|, then \|x\|^2=\|Tx\|^2, for all x. Replace x by x+y and obtain:$$ \langle x,x\rangle+2\langle x,y\rangle+\langle y,y\rangle=\langle ...

2

Pure $\lambda$-calculus is rather restrictive. Nevertheless, you can extend it to contain "$3$" and "$+$" (either by really extending the language, or perhaps by denoting by $3$ some special lambda term like $\lambda f.\ \lambda x.\ f(f(f x))$). Variables are unique symbols (most often strings), which make it easier to construct lambda-terms and define the ...

2

Yes, a sequence is one of its own subsequence, which is similar to an idea that any set is a subset of itself. If we want to specify that a subsequence is not the entire sequence, we refer to it as a proper subsequence. That is, every subsequence except for the sequence itself is a proper subsequence. In the same vein, a proper subset of a set is a ...

2

A point is a primitive notion. In particular, a primitive notion is not defined in terms of previously defined concepts, but is only motivated informally, usually by an appeal to intuition and everyday experience. A vector or a tensor is not an undefined notion, in fact their existence relies on the existence (the definition) of a vector space (tensor ...

2

A sequence can be seen as an ordered list and is typically considered countable, especially in real analysis/calculus, but there are uncountable lists. I would bet that you professor means countable list (ordered by the natural numbers), but one can have "sequences" that are longer than the natural numbers, or even uncountable (I put quotes since sequences ...

2

First let me set the definitions, so there will be no fuss about the terms I use later on. Finite means bijectible with a finite ordinal. Dedekind-finite means "every self-injection is a bijection" (or equivalently, every self-injection is surjective). Tarski-finite means that for every non-empty set of subsets has a $\subseteq$-maximal element. Whenever ...

1

Question 1 You need to be slightly more careful about what you can and can't do. Def. 1 doesn't allow you to conclude that $\frac{a}{a}=1$, only that $\frac{a}{a}=\frac{1}{1}$, and so your proof doesn't work. You won't be able to derive the definition of a fraction from the definition of multiplication. (Indeed, if you have no rule telling you when things ...

1

The interval $(0,1)$ is an ordered, uncountable set in $\mathbb{R}$. It is perhaps slightly tighter to restate his definition as a map from the natural numbers $\{1,2,3,\dots\}$ to the set in question, $\mathbb{R}$ in this case. So while there are uncountable ordered sets in $\mathbb{R}$, they are not sequences since there are not enough natural numbers to ...

1

No. This doesn't work. You have to check that $E_i \cap \left(E_1+E_2+\cdots+E_{i-1}+E_{i+1}+\cdots+E_p\right) = \{0\}$ for each $i$. Just checking that the subspaces don't pairwise intersect is not enough. Consider the example: $U = \{ (0,y) \;|\; y \in \mathbb{R} \}$, $V =\{ (x,0)\;|\; x \in \mathbb{R} \}$, and $W = \{ (x,x) \;|\; x \in \mathbb{R} \}$. ...

1

$~$«$\displaystyle\sum_{i=1}^k E_i$ is direct $\,\Leftrightarrow\,$ the $E_i$s intersect trivially pairwise »$~$ is true for $V$ iff $k<3$ or $\dim V=1$. Proof exercise: Suffices to consider $k=3$, $\dim V=2$. Show if $V=\langle v,w\rangle$ then $\langle v\rangle,\langle w\rangle,\langle v+w\rangle$ is a counterexample to the claim. Notice this does not ...

1

A sequence $f$ is recursive if there is some function $G:\Bbb R^{k+1}\to \Bbb R$ so that $$f_{n+1}=G(f_n,f_{n-1},...,f_{n-k})$$ Then the sequence is determined by providing initial values $f_i=a_i$ for $i=0,1,...,k-1,k$. So no, your definition is strange and seems to define a sequence that is constant for $n>k$. What exactly is the question?

1

First prove that this definition gives a multilinear alternating $n$-form of unit norm on the columns, that is Multilinear (linear in each component): \begin{align} \det (v_1,\dots,v_{i-1},\lambda v_i,v_{i+1},\dots,v_n) &= \lambda\det(v_1,\dots,v_n), \\ \det (v_1,\dots,v_{i-1},v_i+v_i',v_{i+1},\dots,v_n) &= \det(v_1,\dots,v_n) + ...

1

Yes, it is the standard "surface measure" on $S^{n-1}$. In order to prove it, it is enough to check that it gives the expected measure on some simple "surface elements" of your choosing (as long as they generate the Borel algebra). You will find the expression of the spherical volume element (in spherical coordinates) here (we are talking about the standard ...

1

A "space" in mathematics in its most primitive form consists of an underlying set, and elements of this set are usually referred to as "points". But what makes a set an actual "space" is some sort of additional structure imposed on the underlying set, the most simple geometric structure being a topology on the set.

1

You need to define what does it mean "an adequate characterization of finite". If by adequate you mean equivalent in set theory $T$ to a different, presupposed definition of finite (e.g. Dedekind-finiteness), then yes. Taking $T$ as $\sf ZF$ the definition you propose is equivalent to Dedekind-finiteness. To see this, note that your definition is just a ...

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