MA 1124 Solutions 14 th May 2012

MA 1124 Solutions 14 th May 2012 1 (a) Use True/False Tables to prove (i) P = Q Q = P The definition of P = Q is given by P Q P = Q T T T T F F F T T F F T So Q P Q = P F F T T F F F T T T T T Since the P = Q and Q = P columns contain the same truth values in all rows (for all possible truth values of the variables involved), it follows that P = Q Q = P. (ii) (P Q) P Q P Q P Q (P Q) P Q P Q T T T F F F F T F T F F T F F T T F T F F F F F T T T T (iii) (P = Q) (R = Q) (P R) = Q P Q R P = Q R = Q (P = Q) (R = Q) P R (P R) = Q T T T T T T T T T T F T T T F T T F T F F F T F T F F F T T F T F T T T T T F T F T F T T T F T F F T T F T F T F F F T T T F T (b) (i) [ x y (xy = 1)] (where denotes negation) [ x y (xy = 1)] = x [ y (xy = 1)] = x y (xy = 1) = x y xy 1

(ii) [ x y P (x, y) x y Q(x, y)] [ x y P (x, y) x y Q(x, y)] = [ x y P (x, y)] [ x y Q(x, y)] = x y [P (x, y)] x y [Q(x, y)] = x y P (x, y) x y Q (x, y) 2 (a) Two sets X, Y have the same cardinal number means that there exists a bijective mapping f : X Y. [Not same # of elements.] To show this is an equivalence relation, suppose X Y so that there is a bijection f : X Y. Since f is a bijection, f 1 : Y X exists and is also a bijection. Thus Y X. For any X, there exists the identity mapping id X : X X by id X (x) = x. This is clearly bijective so X X. Suppose X Y and Y Z so that there are bijections f : X Y and g : Y Z. But then we proved g f : X Z is bijective, so X Z. Hence is symmetric, reflexive, and transitive. (b) If n is the cardinal number of a set X, and m is the cardinal number of a set Y, then n m means that there exists f : X Y which is 1 1 or injective. Or equivalently X has the same cardinal number as a subset of Y. We can write #X #Y. Since id X : X X is injective #X #X, so is reflexive. If #X #Y and #Y #X, then by the Cantor-Bendixson theorem #X = #Y, so is antisymmetric. If #X #Y and #Y #Z, there are injections f : X Y and g : Y Z. Then g f : X Z is injective, and we have is transitive. Hence is a partial order. (c) (i) (A B) = A B De Morgan s Laws (Note: Venn Diagrams give an illustration, not a proof.) The only way to prove it is as follows. Let x (A B). Then: = x is not in A B = x is not in A and x is not in B = x A and x B = x A B Hence (A B) A B. Now let x A B. Then: = x is not in A and x is not in B = x is not in A B = x (A B) Hence A B (A B). By these two results (A B) = A B.

(ii) A (B C) = (A B) (A C) Let x A (B C). Then: = x A and x (B C) = x A and (x B or x C) = (x A and x B) or (x A and x C) = x (A B) (A C) Let x (A B) (A C). Then: = x (A B) or x (A C) = (x A and x B) or (x A and x C) = x A and (x B or x C) = x A (B C) Hence A (B C) = (A B) (A C). (d) A function f is uniformly continuous on a subset A of R means that ( ) ɛ > 0 δ > 0 c A x A x c < δ = f(x) f(c) < ɛ. (3) (a) Prove lim f(x) = L xn a with x n a f(x n ) L. Given lim f(x) = L and x n a we want to show f(x n ) L i.e. ɛ > 0 N n N with n N f(x n ) L < ɛ. Now lim f(x) = L says that given ɛ > 0, δ such that 0 < x a < δ = f(x) L < ɛ. And x n a, x n a = given δ N such that if n N then 0 < x n a < δ. Putting these together, given ɛ > 0 N such that if n N then f(x n ) L < ɛ. Conversely: to show given x n a, x n a, f(x n ) L, then lim f(x) = L. If you think about this for a while, there is no obvious way to go directly from knowing about all sequences to the limit statement. So we try to assume lim f(x) L, and try to construct x n a such that f(x n ) L. Now lim f(x) L, means ɛ such that δ, x with 0 < x a < δ but f(x) L ɛ. For this ɛ, take δ = 1/n, then x n such that 0 < x n a < δ n. But f(x n) L ɛ. Since 0 < x n a < 1/n, we have x n a and x n a. But f(x n ) L ɛ = f(x n ) L. This is ouer contradiction.

(b) Let {a n } be a sequence. lim inf {a n } = lim k inf {a n} n k. Note if G k = inf {a n } n k, then G k, so if this sequence is bounded lim G k exists. lim sup {a n } = lim k sup {a n} n k. Note if L k = sup {a n } n k, then L k, so if this sequence is bounded lim L k exists. We have G n L m n, m so lim inf {a n } lim sup {a n } and lim inf {a n } = lim sup {a n } lim {a n } exists. If lim {a n } exists then lim sup {a n } = lim inf {a n }. (c) Ā, the closure of A = {x : x n A with x n x} A, the interior of A = {x : ɛ with N(x, ɛ) A} To show (Ā) = (A ) let x (Ā). Then x / Ā x n A such that x n x N(x, ɛ) such that N(x, ɛ) A = (Note if N(x, 1/n) A all n, then we get a sequence x n x with x n A.) But N(x, ɛ) A = = N(x, ɛ) A. Hence x (A ). Hence (Ā) (A ). Conversely, let x (A ). Then = N(x, ɛ) A = N(x, ɛ) A = = x n A such that x n x = x n / A = x (Ā) Hence (A ) (Ā) Hence (Ā) = (A ). (d) Prove that a function from R to R is continuous the inverse image of every open set is open, i.e. Assume f is continuous. Then a ɛ > 0, δ > 0 such that x a < δ = f(x) f(a) < ɛ. Let O be open to show f 1 (O) = {x : f(x) O} is open. (Note: f 1 (A) is always defined for a set A. This is not the inverse function f 1 (y) which may or may not exist.) Let a f 1 (O), then f(a) O. Since O is open, ɛ > 0 such that N(f(a), ɛ) O. But f is continuous at a, so given this ɛ, δ > 0 such that x a < δ = f(x) f(a) < ɛ, i.e. x N(a, δ) = f(x) N(f(a), ɛ). This last is contained in O, so N(a, δ) f 1 (O). Hence a is an interior point of f 1 (O). This is true for all a, so f 1 (O) is open. 4 (a) (i) Let A R be closed and bounded. Prove that every open cover of A has a finite subcover. This means that if A O γ, O γ open, all γ, then we can find a finite number of these, say O γ1, O γ2,..., O γn, such that A O γ1 O γ2 O γn.

Case 1: A an interval, A = [a, b] say. Consider [a, a+b 2 ] O γ, and [ a+b 2, b] O γ. If both of these have a finite subcover, then so does [a, b] and we are done. If one does not, call it I 1. Split I 1 in two halves. If both halves have a finite subcover, so does I 1. So at least one does not. Call it I 2. Keep going. We get a sequence of closed intervals I 1 I 2 I n. Non of these has a finite cover. By the Nested Intervals Property I n. Let x I n. Then x A, so O γ0 such that x O γ0 and O γ0 is open, so N(x, ɛ) O γ0 some ɛ, and x I n, all n, then if the length of I n < ɛ, we have I n N(x, ɛ) O γ0. Here is the picture. I n x ɛ x x + ɛ But then I n has a finite subcover just O γ0 and that intervals as in the last one so don t get them confused. A closed, bounded = A [a, b]. So [a, b] contains all the infinite number of points in our sequence. Consider [a, a+b a+b 2 ], [ 2, b]. One of these must contain an infinite number of terms of the sequence. Call it I 1. Split I 1 into two halves. Again, one of the halves must contain an infinite number of terms of the sequence. Keep going. We get I 1 I 2 I n and each I n contains an infinite number of terms of the sequence. These I n are closed intervals contradicts our choice of all the I n s. This shows that A = [a, b] must have a finite subcover. Case 2: A closed bounded. Then [c, d] with A [c, d]. Suppose A O γ, then [c, d] A ( O γ ) and A is open. Then [c, d] has a finite subcover, say [c, d] A O γ1 O γ2 O γn. But then A O γ1 O γ2 O γn since A is not needed to cover A. (ii) A R, A closed bounded, then every sequence of elements in A has a subsequence that converges to an element of A. Note: this features exactly the same sort of splitting so by the Finite Nested Interval Property, I n. Let x I n and for each I n, choose x n I n to be one of the terms of the sequence. Then x n A, and since the length of I n 0, x n x and A is closed, so x A. (b) A sequence {x n } R is a Cauchy sequence means ɛ > 0 N such that n, m N = x n x m < ɛ. Prove that every Cauchy sequence in R converges to a point in R. Every Cauchy sequence is bounded. Hence {x n } [a, b] some [a, b]. But then {x n } has a convergent subsequence by (a) (ii). But if ** a Cauchy sequence has a subsequence that converges, then the whole sequence converges to that point. We will prove *. Let ɛ = 1, N such that n, m N = x n x m < 1. In particular n N = x n x N < 1. So all x n, n 1 lie in the interval (x N 1, x N + 1). Hence {x n } are bounded above by max {x 1, x 2,..., x N, x N + 1} and below by

min {x 1, x 2,..., x N, x N 1}. We will prove **. Given {x n } Cauchy sequence and x nk x. Claim x n x. Given ɛ > 0, n k0 such that n k n k0 = x nk x < ɛ/2 (since x nk x). N such that n, m N = x n x m < ɛ/2 (since x n is a Cauchy sequence). Hence if n N, choose n k1 max {N, n k0 } then x n x x n x nk1 + xnk1 x < ɛ.