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Problem S1: Let Ω be a collection of subsets of a set X. Formulate and prove de Morgan’s laws for the intersection and union of sets from Ω. Solution. Let Ω = {A1 , A2 , . . . , An } n T C n AC S First we prove that Ai = i . i=1 i=1 n T C n T Proof. Let x ∈ Ai ⇔x∈ / Ai ⇔ (x ∈ / A1 ) ∨ (x ∈ / A2 ) ∨ · · · ∨ (x ∈ / An ) i=1 i=1 n / Aj ⇒ x ∈ AC AC S So, for some j, x ∈ j =⇒ x ∈ i . i=1 T n  C n AC S So, Ai ⊆ i . (∗) i=1 i=1 n AC C S Let x ∈ i ⇔ x ∈ Aj for some j. i=1 n T Tn C So, x ∈ / Aj ⇒ x ∈/ Ai =⇒ x ∈ Ai i=1 i=1 Tn C n C S So, Ai ⊇ Ai . (∗∗) i=1 i=1 n T C n AC S (∗) with (∗∗) implies Ai = i .  i=1 i=1 n S C n AC T Next we prove that Ai = i . i=1 i=1 n S C n S Proof. Let x ∈ Ai ⇔x∈ / Ai ⇔ (x ∈ / A1 ) ∧ (x ∈ / A2 ) ∧ · · · ∧ (x ∈ / An ) i=1 i=1 n / Aj ⇒ x ∈ AC AC T So, for all j, x ∈ j =⇒ x ∈ i . i=1 S n  C n AC T So, Ai ⊆ i . (∗) i=1 i=1 n AC T Let x ∈ i ⇔ (x ∈ / A1 ) ∧ (x ∈ / A2 ) ∧ · · · ∧ (x ∈ / An ) i=1 Sn  n S C So, for all j, x ∈ / Aj ⇒ x ∈ / Ai =⇒ x ∈ Ai . i=1 i=1 n S C n AC T So, Ai ⊇ i . (∗∗) i=1 i=1 n S C n AC T (∗) with (∗∗) implies Ai = i .  i=1 i=1 1 2 Problem S2: Let A and B be subsets of R and let A + B be the set of all numbers of the form a + b, where a ∈ A ⊂ R and b ∈ B ⊂ R. Prove that sup(A + B) = supA + supB. Solution. By definition of supremum: supA = x ≥ a for any a ∈ A (∗) supB = y ≥ b for any b ∈ B (∗∗) So, if we consider any a ∈ A, b ∈ B then from (∗) and (∗∗): x+y ≥a+b∈A+B and by definition of supremum, sup(A + B) = x + y = supA + supB.  Problem S3: Prove the binomial formula n X (a + b)n = Cnk an−k bk . i=1 n z }| { Solution. Consider the expression (a + b)(a + b) . . . (a + b). There is exactly one a and one b in each bracket, so if we want to choose n a-s and n − n b-s, we can do it in Cnn = Cn0 = 1 way. Analogously, if we want to choose k a-s and n − k b-s, we can do it in Cnk = Cnn−k ways. So, from the all above-mentioned we get: n X (a + b)n = Cnn an + Cnn−1 an−1 b + · · · + Cnn−1 abn−1 + Cn0 bn = Cnk an−k bk .  i=1 3 Problem S4: Is the following theorem true: If lim g(x) = A and lim f (x) = B then lim f (g(x)) = B x→a x→A x→a Solution. No. Proof. Let ( sin(x) , if x 6= 0 f (x) = x g(x) = 0 0, if x = 0 Then lim g(x) = 0, lim f (x) = 1 and lim f (g(x)) 6= 1.  x→0 x→0 x→0 Problem S5: Let I ⊂ R be an interval and let f : I → R be continuous. Prove that f is injective (into) if and only if it is strictly monotonous (i.e. either strictly monotonously increasing or decreasing). Solution. Let f be injective but not strictly monotonous. By definition of in- jectivity ∀a, b ∈ I f (a) = f (b) ⇒ a = b. If f has local maxima, then ∃x, y, z ∈ I : x < y < z (f (x) < f (y)) ∧ (f (y) > f (z)). Without loss of generality let f (x) < f (z) ⇒ f (x) < f (z) < f (y). By intermediate value theorem ∃t ∈ (x, y) : f (t) = f (z). Contradiction with injectivity of f . Case with local minima is similar. Let f be strictly monotonous but not injective. ∃x, y ∈ I : (x 6= y) ∧ (f (x) = f (y)) Contradiction with strictly monotone.  4 Problem S6: By utilizing the ε − δ-definition of limits show that a) the function f : R → R is continuous in x0 = −1, where x−1 f (x) = 2 . x +1 b) the characteristic function χ : R → {0, 1} of the rational numbers is not contin- uous at any point x ∈ R, where  1, if x ∈ Q χ(x) = 0, if else. Solution. a) We need to proove that ∀ε > 0 ∃δ > 0 ∀x : 0 <| x + 1 |< δ ⇒| f (x) − f (−1) |< ε x−1 x2 + 1  1  | f (x)−f (−1) |< ε ⇔ + 1 < ε ⇔ |x+1| < ε ⇔ |x+1| < |x|+ ε x2 + 1 |x| |x| 1 3 1 1 5 5 if |x + 1| < then − < x < − ⇒ |x| + < ⇔ |x + 1| < ε 2 2 2 |x| 2 2 So, δ =min{ 21 , 52 ε} works. b) First we prove two statements: Theorem 1. Between any two distinct real numbers there is a rational number. Proof. Let x, y ∈ R, x < y. We choose n ∈ N such that n1 < y − x, consider the sequence n1 , n2 , n3 , . . . and choose the largest k ∈ N such that nk ≤ x. Then k+1 n >x and n1 = k+1 n − k n < y − x ⇒ k+1 n < y. So, we found rational number k+1 n between x and y.  Theorem 2. Between any two distinct rational numbers there is a irrational num- ber. Proof. Let √ x, y ∈ Q, x < y. 2(y−x) z =x+ 2 is a irrational number between x and y.  Now we can prove b). Consider any irrational number x0 . χ(x0 ) = 0, consider 0 < ε < 1, then by Theorem 1 for any δ > 0 we can find a rational number x1 from the interval (x0 − 2δ , x0 + 2δ ). By definition of χ(x), χ(x1 ) = 1, so lim χ(x) x→x0 doesn’t exists. Consider any x0 ∈ Q. χ(x0 ) = 1, consider 0 < ε < 1, then by Theorem 2 for any 0 < δ ∈ Q we can find a irrational number x1 from the interval (x0 − 2δ , x0 + 2δ ). Also, for any irrational δ > 0 we can find a irrational number x1 = x0 ± 4δ . By definition of χ(x), χ(x1 ) = 0, so lim χ(x) doesn’t exists. x→x0 This proves that χ(x) is not continuous at any point x ∈ R  5 Problem S7: Let n ∈ N and Nn = {1, 2, ..., n}. Prove that for all functions f : Nn → Nm with n > m there are two different numbers n1 , n2 ∈ Nn such that f (n1 ) = f (n2 ). Solution. Consider pair for every value of function f : (f (1), 1), (f (2), 2), . . . , (f (m), m)) n > m, so there is at least one ni ∈ Nn without pair. So, for some ni , nj ∈ Nn f (ni ) = f (nj ).  Problem S8: Let p(x) be an arbitrary polynomial of degree three. a) Determine limx→+∞ p(x) and limx→−∞ p(x) as well as the supremum and infimum of p(x). b) Apply the intermediate value theorem to show that p(x) has at least one real root. Solution. Let p(x) = ax3 + bx2 + cx + d. b c d a) lim p(x) = lim ax3 +bx2 +cx+d = lim x3 × lim a+ + 2 + 3 = +∞ (∗) x→+∞ x→+∞ x→+∞ x→+∞ x x x b c d lim p(x) = lim ax3 +bx2 +cx+d = lim x3 × lim a+ + 2 + 3 = −∞ (∗∗) x→−∞ x→−∞ x→−∞ x→−∞ x x x From (∗) and (∗∗) we can see that p(x) doesn’t have supremum or infimum. b) From (∗) and (∗∗) we get that for enough big/small x, p(x) > 0 and for enough small/big x, p(x) < 0. So, by intermediate value theorem p(x) has at least one root.  Problem S9c: Evaluate the limit √ m x−1 lim √ x→1 n x−1 Here m, n ∈ N. √ √ Solution. Let x = tmn ⇒ m x = tn and n x = tm . x → 1 ⇒ xmn → 1 ⇒ t → 1 √ m x−1 tn − 1 (t − 1)(tn−1 + tn−2 + · · · + t + 1) lim √ = lim = lim = x→1 n x − 1 x→1 tm − 1 x→1 (t − 1)(tm−1 + tm−2 + · · · + t + 1) (tn−1 + tn−2 + · · · + t + 1) n = lim = x→1 (tm−1 + tm−2 + · · · + t + 1) m 6 Problem S10: Give the (maximal) domain and (maximal) range of  1  x 7→ f (x) := exp cos(x) and determine the left-hand and right-hand limits for all boundary points of the domain. Solution. π cos(x) 6= 0 ⇒ x 6= + πk, k ∈ Z 2 π D(f ) = R \ { + πk, k ∈ Z}. 2  1  1 1 y = exp ⇒ lny = ⇒ cos(x) = cos(x) cos(x) lny 1 −1 ≤ ≤1 lny 1 ≤ 1 ⇒ y ∈ (0, 1) ∪ [e, +∞) lny 1  1i ≥ −1 ⇒ y ∈ 0, ∪ [1, +∞) lny e  1i So, y ∈ 0, ∪ [e, +∞) e  1i E(f ) = 0, ∪ [e, +∞). e  1   1  lim exp = +∞ lim exp =0 x→ π 2 − cos(x) x→ π 2 + cos(x)  1   1  lim exp = 0 lim exp = +∞ x→ 3π2 − cos(x) x→ 3π2 + cos(x) Since f is periodic with T = 2π :  1   1  lim exp = +∞ lim exp =0 x→( π 2 +2πk) − cos(x) x→( π2 +2πk) + cos(x)  1   1  lim exp =0 lim exp = +∞. x→( 3π 2 +2πk) − cos(x) x→( 3π 2 +2πk) + cos(x)