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17 Nov 2013: Use (ii) to show that there exists b K such that det(σiσjg1(b)) 6= 0.Deduce that if y = g1(b) then {σ(y) | σ G} is a K-basis ... for L.Such a basis {σ(y)} is said to be a normal basis for L/K, and the result just proved is the NormalBasis Theorem.

12 Mar 2013: Show that I T hasa square root, i.e., there exists S B(X) such that S2 = I T. ... invertible operator T : X X such that Txn = yn for all n N.

27 Nov 2013: 6. Let a1, a2,. , an be real numbers such that a1 an = 0 and a21 a. ... 9. (i) Show that if x 6= y are vectors in the finite dimensional vector space V , then there is a linearfunctional θ V such that θ(x) 6=

31 Jan 2013: 16. Find X R2 such that (a) any two points x,y X can be joined by a continuouspath γ : [0, 1] X and (b) for x 6= y the length ... of any such path is infinite.

3 Nov 2013: Find a, b, c Q such thaty = a bx cx2. ... 15. Let K and L be subfields of a field M such that M/K is finite.

27 Nov 2013: 12. Let S be a collection of subsets of N such that for every A, B S we have A B orB A. ... otherwords, for every a < b and every c there is an x with a < x < b such that f(x) = c.

17 Nov 2013: Use (ii) to show that there exists b K such that det(σiσjg1(b)) 6= 0.Deduce that if y = g1(b) then {σ(y) | σ G} is a K-basis ... for L.Such a basis {σ(y)} is said to be a normal basis for L/K, and the result just proved is the NormalBasis Theorem.

23 Oct 2013: 11 fn 6 0. Is it possible to find such a sequence with |fn(x)| 1 for all n and for all x? ... On the other hand, show that if fn does notconverge uniformly to f then we can find a convergent sequence xm x in [a,b] such that fn(xn) 6

3 Nov 2013: Find a, b, c Q such thaty = a bx cx2. ... 15. Let K and L be subfields of a field M such that M/K is finite.

7 May 2013: Remark 2.2. — There exist coefficient systems B M such that. ... Let β H(M; (m )B) be a class such that (m )(β) = 0 H(M; B).

22 Feb 2013: Show that for any b C, there exists a sequence of points zn Ba(r) withzn 6= a such that zn a and f(zn) b as n. ... Find such a sequence when f(z) = e1/z, a = 0 and b = 2.

16 Oct 2013: 1. Calculate d = (a,b) and find integers x and y such that d = ax by when. ... If n is a prime,deduce that there are infinitely many primes q such that q 1(mod n).

2 Dec 2013: 6. Let a1, a2,. , an be real numbers such that a1 an = 0 and a21 a. ... 9. (i) Show that if x 6= y are vectors in the finite dimensional vector space V , then there is a linearfunctional θ V such that θ(x) 6=

7 Jan 2013: Show that there is a variable length coding. c such that c is injective and all codewords have length 2 or less. ... Show that there is nodecodable coding c such that all codewords have length 2 or less.

5 Feb 2013: 16. Find X R2 such that (a) any two points x,y X can be joined by a continuouspath γ : [0, 1] X and (b) for x 6= y the length ... of any such path is infinite.

4 Dec 2013: Show that I Thas a square root, i.e., there exists S B(X) such that S2 = I T. ... 14+. Let X be a separable infinite-dimensional Banach space. Suppose thatthere is a constant C such that every finite-dimensional subspace E of X satisfiesd(E,n2 ) C, where

16 Oct 2013: C(M̊,X) (C((0, 1) Sd1,X))• C(Rd,X). is such a semi-simplicial object via the H-space and H-module structure maps (andthe simplicial identities hold by ... x, of bidegree (k(d 1),m), is such that(3.4) x [n] = A k [n mk] B k1 τ [n mk 1]for n 0, then if

13 Aug 2013: possibly with boundary,and where global coordinates u and v can be chosen on Q such that. ... Hence such surfaces must be absent. To prove Theorem 1.1 we need to establish the stronger statement that no hori-zons will form.

12 Sep 2013: emαm! preserves M. (b) For all b U(g) there exists i > 1 such that ad(reα)i. ... Then thereexists b U(g)n such that a b mod πNÛ(g)n. Now D.

20 Feb 2013: Then thereexists a RACSDC automorphic representation π of GLn(AE) such that Π = πL. ... and moreover such that there is an isomorphism. rιp(π′)|IFu =. (ψ 00 ψqu. ),.

3 Apr 2013: On the other hand, the stabilizerof such a pair in G(K) is trivial. ... pdr1,x,y) on X such that the family X B of curves is asgiven by the following table:.

21 Apr 2013: Still assume that R Zp. There is a unique γp 1 pZp such thatγmp = δ. ... a(n)qn/m =nZ. b(n)tn, b(n) o. Let H =rj=0 AjT. j o[T] such that the image of f in DR(Γ,o,k) is

19 Jun 2013: p])representing ξ and γ R such that all of the following conditions are satisfied. ... A known to be isomorphic to Mat3(K), find such an isomorphism explicitly.

15 Mar 2013: underlying 1-cell f : a P b we give a pseudomap with underlying1-cell f : P a P b, in such a way that the 2-cells ηf , κA and ... 23. and there is a free algebraic theory generated by such an operad.

20 Feb 2013: Such a group exists since [F : Q] is even, and is uniquely determined upto isomorphism. ... OB OF OFv = Mn(OEw ) Mn(OEwc ),. such that † acts as (g1,g2) 7 (g2, tg1).

25 Feb 2013: Such g is called a local inverse of f. The degree of holomorphic map. ... The existence of such Ṽ is a consequence of thecompactness of R and the continuity of f.

18 Dec 2013: See other footnotes below. 7We can now give such Lorentzian manifolds a name: they are called time-orientable. ... A Killing field is a vector field V on M such that LLg = 0; here L denotesLie-differentiation.

13 Jan 2013: vm,v are linearly dependent, then there exist ci,b such thatcivi bv = 0, with not all ci,b zero. ... 0 0. that is, there exist invertible P GLm(F), Q GLn(F) such that B =Q1AP.