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19 Dec 2003: Show that iγi(D). B(p̃, 4d).(c) Using that the sets γi(D) are disjoint and all have the same volume show that. ... a) Prove that B(p,δ) contains D for all p D.(b) Show that.

5 Nov 2003: It is the set of all elements of A that are not elements of B. ... AxB, the Cartesian product of A and B, is defined to be the set of all ordered pairs (x,y) such that x is an element of A and y is

26 Jun 2003: replacemin computes a tree in which every leaf ofthe argument has been replaced by the minimum of all theleaves. ... Definition 2.1. A morphism f : A B in a premonoidalcategory K is central if for all g : A′ B′ in K, f g = f g.If at least one of

19 May 2003: α(λ1u1 λ2u2) = λ1α(u1) λ2α(u2). for all uj Uj, λj F. ... Then () hasa unique solution for all b if and only if detA 6= 0.

24 Nov 2003: x < y.]. 6. (i) Let Q(. 2) be the subset of R consisting of all numbers of the form a b. ... 9. Let r be a real number and let A be the set of all rational numbers less than r.

17 Jul 2003: In all these examples|A| = 3 and B =. According to mwrank, rank Dλ(Q) = 0. ... θ (Qp )n for all p B}. We have already seen two proofs of the statement concerning torsion, in Sect.

19 Dec 2003: Show that for all t, Kγ̇(t)(ei(t)) =λi ei(t), where λi is independent of t. ... Show that all the geodesics of S3/G are closed,but they could have different lengths.

29 Sep 2003: for all x, otherwise d(x,y) = 3n if xy = 3na/b where a,b are prime. ... iii) Let φ be a continuous real-valued function on R[a,b] which satisfies the Lipschitz condition|φ(x,t) φ(y,t)| M |xy| , for all x,y R

9 Jun 2003: For example, one could divide a class into the set G of all girls and the set B of all boys. ... To give a simple example: if I take two subsets A and B, I can define their union to be the set of all n such that n belongs either to

7 Oct 2003: Note that if Cis a lccc then so are all its slices: for an object A in C and an object f : B Ain C/A, there is indeed an isomorphism ... iii) The endofunctor PX has an initial algebra, for all X in C.

2 Aug 2003: nk=0(t xk) which is small for all t [a,b]. Such a polynomial was. ... In other words, we can findx1, x2 [a,b] such that. f(x1) f(x) f(x2) for all x [a,b].

14 May 2003: Everything takes place inside the vector space of all real-valued functions on B. ... ij. =a. i. b. j. Notice that matrices of the form v@w are only a small subset of all m-by-n matrices.

21 Aug 2003: leaf of the argument has been replacedby the minimum of all the leaves. ... A morphism f : A B in a premonoidal category K is centralif for all g : A′ B′ in K, f g = f g.

3 Sep 2003: S. That sounds pretty convincing. So let's include all algebraic numbers then. ... All right. It says that if f is a continuous function, a< b are real numbers and f(a)=u, f(b)=v, then for any w between u and v

10 Sep 2003: hypothesis (1) translates into:. LΘ(c) > 0, for all c BC,and hence the closed subspace Ker(LΘ) B meets the cone C of foliation currentsonly at zero. ... P. PATERNAIN. • F(c) > 0 for all 0 6= c C.Let us consider the restriction of F and LΘ to B.

19 Feb 2003: ALc(γ|[s,t]. )= Φc(γ(s),γ(t)). for all a s t b. Semistatic curves are solutions of the Euler–Lagrangeequation because of their minimizing properties. ... Φc(γ(s),γ(t)) Φc(γ(t),γ(s)) = 0. for all a s t b. The notions of static and semistatic

11 Mar 2003: symplectically invariant.The right hand side is always between 0 and n for all t. ... Then. (a) Π(e(Σ)) has no conjugate points.(b) e(z) E(Π e(z)) = {0} for all z Σ.