State which of the following statements are true or false, in either case substantiate.3.1 Let S = (1, 2, 3) and r be a topology on S defined by r = {0, {1}, {2), (2,3), S). Then(a) (S, T) is a normal space,(b) (S, T) is a regular space,(c) (S, T) is not a T₁-space,(d) (1) and (2, 3) are separated sets,3.2 The set Q of rational numbers as subset of the set R of real numbers with discretetopology is open.Page 3 of 3

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Answered: 3.1 Let S = (1,2,3) and be a topology…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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6. Suppose V is a real inner product space. (a) Show that if u, v € V with ||u|| = ||v||, then u + v is orthogonal to u – v. (b) Provide an example of a real inner product space V and u, v Є V such that u + v is not orthogonal to u - v.
(b) Let X bę a normed linear space. Let Вх, г) %3D (у є Х: || x-у ||
topology
Once you get away from vector spaces of finite dimension, it is no longer true that every linear map is automatically continuous. For example, let Coo {f E F(N, R) : f(n) = 0 for all sufficiently large n}, where F(N, R) is the vector space of all func- tions from N to R with the usual pointwise operations. Another way of expressing the condition for f to belong to Coo is that {n : f(n) # 0} is finite. Then define L: C00R by L(f) = Enf(n). Notice that the sum on the right hand side is in fact a finite sum. Now (a) Check that L is linear. (b) Show that {|L(f)| : ||F|| < 1} is not bounded above. Then deduce that L is not continuous, from the following theorem: For each continuous linear map L:V W the set {I|L(x)|lw: x|lv <1} is bounded above. 1 n=r (Hint: consider the functions e,, defined by e, (n) = and define a %3D 0 otherwise norm on e, by ||e,|| := sup{e,(n)|: n < 1} = le,(1)|. Show that for all r E N, |le,|| < 1. On the other hand {|L(e,)|:r€ N} is not bounded above.]
Q3) Let L = Z* and |x ||= max {|x1 - x3l, x2l, 15x,1} Vx = (x1, X2, X3, X4) E Z*. Is (Z*, ) is a normed space?
1) Let V, W be F vector spaces. Recall that a skew-symmetric map is a function f: V × V → W such that ƒ(V₁, V2) = −ƒ (V₂, V₁) (a) (b) Prove that an alternating map is always skew-symmetric Prove that if char(F) + 2 then a skew-symmetric map is also alternating
2. Let V, W be subsets of R“ defined the following way: a а 2а + b a+b V = a,b ER W : a,b €R, a<0
2. Check that the following are not vector spaces by proving that at least one axiom in the definition of a vector space is false. (a) The quadruple (R, R", H, O2). Here, H denotes the usual operation of addition multiplication among vectors, while O2 denotes the nonstandard operation of scalar multiplication given by c(x1, x2, . .. , Xn) = (cx1,0, 0, . . .,0). (b) The quadruple (R, V, H, D), where V denotes in this case the subset of Rmxn whose elements have the entry in row 1 and column 1 equal to some r E R, r + 0. Here, B and O denote the usual operations of addition and scalar multiplication among matrices.
4. Let V = {ao+aj x+a2 x² | ao, a1, a2 E R} be a vector space of polynomial up to degree three, and p(x) = pPo+P1 x+p2 x², q(x) = qo+q1 x+q2 x² E V then show that = po.4o + pP1-q1 + P2-92 defines an inner product on V.
14. If d is a metric on a vector space X‡ {0} which is obtained from a norm, and å is defined by ã(x, x) = 0, ã(x, y) = d(x, y)+1 (x‡y), show that a cannot be obtained from a norm.
Show that (5,0,3) is in the span of S= {(1,1,0), (1,0,1), (1,1,1)}.
. If V is an inner product space, then for any vectors a, ß in V and any scalar c (i) l|ca|| = |c||la||; (ii) |la|| > 0 for a + 0; (iii) (ælß)| < ||a|| ||ß|l; (iv) ||a + B|| < lla||+ I|B|I.

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