CMI Entrance Exam 2020 Question Paper Solution M.Sc Maths

CHENNAI MATHEMATICAL INSTITUTE Postgraduate Programme in Mathematics MSc/PhD Entrance Examination 4th October 2020 Part A (1) A, B. (2) A, C. (3) A. (4) D. (5) A. (6) C, D. (7) A, C, D. (8) B. (9) A, D. (10) 3. Part B (11) It is easily checked that δ is a metric. Let Ck be a sequence of compact subsets of X. We must show that there is a convergent subsequence. Given any  > 0, using there exists an N = N () such that the every Ck is covered by (atmost) N () many open balls of radius  with centres in Ck . We do this first for X with /2 balls and then take, for each such ball B, an -ball about a point in Ck in case the intersection B ∩Ck is non-empty. We let  vary over 1/n, n ∈ N. This way we get a finite set Fk,n ⊂ Ck of centres of balls used to cover Ck . We may assume that Fk,n ⊂ Fk,n+1 by including P balls of radius (n+1) around points of Fk,n . The cardinality of Fk,n is at most m≤n N (1/m) =: Nn which is independent of k.) Then δ(Fk,n , Ck ) < 1/n. We list the points of ∪n≥1 Fk,n in a sequence: xk,1 , xk,2 , . . . , xk,j , . . . , viewed as kth row of a matrix whose rows and columns are labelled by N. It is understood that in the above sequence we list members of Fk,i before those of Fk,i+1 ∀i ≥ 1 for each k ≥ 1. We consider the first column. Now xk,1 , k ∈ N is a sequence of points in X which has a convergent subsequence xkr ,1 ∈ Ckr , r ∈ N . Set y1 to be the limit of this subsequence. We will denote Ck1 as C11 . We consider second column entries corresponding to the rows labelled by the sub- sequence kr —that is the sequence xkr ,2 . This has a convergent subsequence, say with limit y2 . We will denote the first term of the corresponding subsequence by C22 . Proceeding thus we obtain a sequence of points y1 , . . . , yn , . . .. We let C be the closure of {yk | k ≥ 1}. Then C is compact. Our claim is that the ”diagonal” sequence Ckk converges to C. Let n > 0 be a positive integer. Given any m ∈ N we have δ(Cm , Fm,n ) < 1/n and so δ(xm,r , Fm,n ) < 1/n for all r. P So yr is at a distance at most 1/n from Fn := {y1 , y2 , . . . , ym | m ≤ Nn } where Nn = l≤n N (1/l). So it follows that δ(C, Fn ) ≤ 1/n. If Ckk = Cr , (r = r(k) depends on k) for sufficiently large k we have δ(xr,j , yj ) < 1/n for j ≤ Nn . Hence we have δ(Fr,n , Fn ) < 1/n. So δ(Cr , C) ≤ δ(Cr , Fr,n ) + δ(Fr,n , Fn ) + δ(C, Fn ) < 3/n. (12) Let C be a component.Then C is open. If C is bounded, the value of |f | on its boundary ∂C is 1 ( by continuity, open mapping theorem and because ∂C is disjoint with C). Hence same for 1/|f |. But this contradicts |f (z)| < 1 on C. 1

1 (13) Let f : X −→ R be given by x 7→ min{d(x,p)|p∈F } . It is continuous. The graph Γf := {(x, f (x)) | x ∈ X} is a closed subset of X × R, hence a complete metric space. X 7→ Γf , x 7→ (x, f (x)) is a homeomorphism. (14) Note that f [0, ∞) ⊂ [0, ∞) as f 0 ≥ 0. Integrating f 0 /f 2 on [0, x] gives 1 − 1/f (x) ≥ x and hence f is unbounded in [0, P 1). (15) Let c1 , . . . , cn ∈ C be such that i ci eai z = 0. Differentiating this n − 1 times, we see that i ci aji eai z = 0 for every 0 ≤ j ≤ n − 1. Substitute z = 0 to get i ci aji = 0 for P P every 0 ≤ j ≤ n − 1. The Vandermonde matrix (aji )i,j is invertible, so ci = 0 for every i. (16) contain m for any integer m with gcd(m, pn ) = 1, so H = {jp | 0 ≤ j < p}. (17∗ ) If M is diagonalizable, then we may assume that M is a diagonal matrix. Then P (M ) is a diagonal matrix. Such a matrix is nilpotent if and onlyQif it is zero. Conversely, let α1 , . . . , αm be the distinct eigenvalues of M . Let P (X) = m i=1 (X − αi ). Let µ(X) be the minimal polynomial of M . Since the roots of µ(X) are exactly α1 , . . . , αm , there exists a positive integer r such that (P (X))r is divisible by µ(X). Hence (P (M ))r = 0. Therefore P (M ) = 0, i.e., P (X) = µ(X). Therefore M is diagonalizable. (18∗ ) (1) True: In fact, X is path connected. We know the closed interval [0,1] is normal. Given two points x, y ∈ X, consider the function f : {0, 1} −→ X defined by f (0) = x, f (1) = y. Then f is continuous and extends to all of [0, 1], by the hypothesis on X. (2) False: take a finite set of cardinality at least 2. Then it is compact and not connected. So it can’t have universal extension property. (3) Note that X ⊂ R2 is a retract: r : R2 −→ X given by f (x, y) = (x, sin x) is a retraction. Since R2 has universal extension property, any map A −→ R2 can be extended to all of Y (for any given pair of a normal space Y and a closed subset A ⊂ Y ). If we are given a function f : A −→ X, composing with the inclusion X −→ R2 , we have an extension g : Y −→ R2 . Composing this with r, we get the desired extension Y −→ X. Note that it is not required to know about the language of retracts for (3). They will have to notice that a continuous map like r exists. (19∗ ) If f (a) = f (b) = 0 for some a, b ∈ K, then so f (λa + b) = 0 for every λ ∈ Fq . Hence the set of zeros is an Fq -vector-space. It is n-dimensional since f is separable. (20∗ ) First n(n + 1) and n2 are of the same order (the ratio converges to one). So you can as well take denominator as n(n + 1). Thus need to show a1 + 2a2 + 3a3 + · · · + nan → a. n(n + 1)/2 Imitate blindly Cesaro theorem. Given  > 0, choose K so that |an − a| < /2 for n ≥ K. Then choose N > K so that the finite sum |a1 − a| + 2|a2 − a| + 3|a3 − a| + · · · + K|aK − a| < /2. N (N + 1)/2 If now n > N then a1 + 2a2 + 3a3 + · · · + nan |bn − a| = | − a| n(n + 1)/2 (Use 1 + · · · + n = n(n + 1)/2 to distribute a to each term in numerator.) |a1 − a| + 2|a2 − a| + 3|a3 − a| + · · · + n|an − a| ≤ n(n + 1)/2 (split first K terms and the remaining.)   < + 2 2 2