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 = { A 1 , A 2 , . . . , A n } First we prove that ( n ⋂ i =1 A i ) C = n ⋃ i =1 A C i Proof. Let x ∈ ( n ⋂ i =1 A i ) C ⇔ x / ∈ n ⋂ i =1 A i ⇔ ( x / ∈ A 1 ) ∨ ( x / ∈ A 2 ) ∨ · · · ∨ ( x / ∈ A n ) So, for some j , x / ∈ A j ⇒ x ∈ A C j = ⇒ x ∈ n ⋃ i =1 A C i So, ( n ⋂ i =1 A i ) C ⊆ n ⋃ i =1 A C i ( ∗ ) Let x ∈ n ⋃ i =1 A C i ⇔ x ∈ A C j for some j So, x / ∈ A j ⇒ x / ∈ n ⋂ i =1 A i = ⇒ x ∈ ( n ⋂ i =1 A i ) C So, ( n ⋂ i =1 A i ) C ⊇ n ⋃ i =1 A C i ( ∗∗ ) ( ∗ ) with ( ∗∗ ) implies ( n ⋂ i =1 A i ) C = n ⋃ i =1 A C i Next we prove that ( n ⋃ i =1 A i ) C = n ⋂ i =1 A C i Proof. Let x ∈ ( n ⋃ i =1 A i ) C ⇔ x / ∈ n ⋃ i =1 A i ⇔ ( x / ∈ A 1 ) ∧ ( x / ∈ A 2 ) ∧ · · · ∧ ( x / ∈ A n ) So, for all j , x / ∈ A j ⇒ x ∈ A C j = ⇒ x ∈ n ⋂ i =1 A C i So, ( n ⋃ i =1 A i ) C ⊆ n ⋂ i =1 A C i ( ∗ ) Let x ∈ n ⋂ i =1 A C i ⇔ ( x / ∈ A 1 ) ∧ ( x / ∈ A 2 ) ∧ · · · ∧ ( x / ∈ A n ) So, for all j , x / ∈ A j ⇒ x / ∈ n ⋃ i =1 A i = ⇒ x ∈ ( n ⋃ i =1 A i ) C So, ( n ⋃ i =1 A i ) C ⊇ n ⋂ i =1 A C i ( ∗∗ ) ( ∗ ) with ( ∗∗ ) implies ( n ⋃ i =1 A i ) C = n ⋂ i =1 A C i 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 ) = sup A + sup B Solution. By definition of supremum: sup A = x ≥ a for any a ∈ A ( ∗ ) sup B = 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 = sup A + sup B Problem S3: Prove the binomial formula ( a + b ) n = n ∑ i =1 C k n a n − k b k Solution. Consider the expression n ︷︸︸︷ ( 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 C n n = C 0 n = 1 way. Analogously, if we want to choose k a -s and n − k b -s, we can do it in C k n = C n − k n ways. So, from the all above-mentioned we get: ( a + b ) n = C n n a n + C n − 1 n a n − 1 b + · · · + C n − 1 n ab n − 1 + C 0 n b n = n ∑ i =1 C k n a n − k b k 3 Problem S4: Is the following theorem true: If lim x → a g ( x ) = A and lim x → A f ( x ) = B then lim x → a f ( g ( x )) = B Solution. No. Proof. Let f ( x ) = { sin ( x ) x , if x 6 = 0 0 , if x = 0 g ( x ) = 0 Then lim x → 0 g ( x ) = 0, lim x → 0 f ( x ) = 1 and lim x → 0 f ( g ( x )) 6 = 1. 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 x 0 = − 1, where f ( x ) = x − 1 x 2 + 1 b) the characteristic function χ : R → { 0 , 1 } of the rational numbers is not contin- uous at any point x ∈ R , where χ ( x ) = { 1 , if x ∈ Q 0 , if else. Solution. a) We need to proove that ∀ ε > 0 ∃ δ > 0 ∀ x : 0 < | x + 1 | < δ ⇒| f ( x ) − f ( − 1) | < ε | f ( x ) − f ( − 1) | < ε ⇔ ∣ ∣ ∣ ∣ x − 1 x 2 + 1 + 1 ∣ ∣ ∣ ∣ < ε ⇔ | x +1 | < x 2 + 1 | x | ε ⇔ | x +1 | < ( | x | + 1 | x | ) ε if | x + 1 | < 1 2 then − 3 2 < x < − 1 2 ⇒ | x | + 1 | x | < 5 2 ⇔ | x + 1 | < 5 2 ε So, δ =min { 1 2 , 5 2 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 1 n < y − x , consider the sequence 1 n , 2 n , 3 n , . . . and choose the largest k ∈ N such that k n ≤ x . Then k +1 n > x and 1 n = 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 z = x + √ 2( y − x ) 2 is a irrational number between x and y Now we can prove b) . Consider any irrational number x 0 χ ( x 0 ) = 0, consider 0 < ε < 1, then by Theorem 1 for any δ > 0 we can find a rational number x 1 from the interval ( x 0 − δ 2 , x 0 + δ 2 ). By definition of χ ( x ), χ ( x 1 ) = 1, so lim x → x 0 χ ( x ) doesn’t exists. Consider any x 0 ∈ Q χ ( x 0 ) = 1, consider 0 < ε < 1, then by Theorem 2 for any 0 < δ ∈ Q we can find a irrational number x 1 from the interval ( x 0 − δ 2 , x 0 + δ 2 ). Also, for any irrational δ > 0 we can find a irrational number x 1 = x 0 ± δ 4 By definition of χ ( x ), χ ( x 1 ) = 0, so lim x → x 0 χ ( x ) doesn’t exists. This proves that χ ( x ) is not continuous at any point x ∈ R 5 Problem S7: Let n ∈ N and N n = { 1 , 2 , ..., n } Prove that for all functions f : N n → N m with n > m there are two different numbers n 1 , n 2 ∈ N n such that f ( n 1 ) = f ( n 2 ). 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 n i ∈ N n without pair. So, for some n i , n j ∈ N n f ( n i ) = f ( n j ). Problem S8: Let p ( x ) be an arbitrary polynomial of degree three. a) Determine lim x → + ∞ p ( x ) and lim x →−∞ 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 ) = ax 3 + bx 2 + cx + d a) lim x → + ∞ p ( x ) = lim x → + ∞ ax 3 + bx 2 + cx + d = lim x → + ∞ x 3 × lim x → + ∞ a + b x + c x 2 + d x 3 = + ∞ ( ∗ ) lim x →−∞ p ( x ) = lim x →−∞ ax 3 + bx 2 + cx + d = lim x →−∞ x 3 × lim x →−∞ a + b x + c x 2 + d x 3 = −∞ ( ∗∗ ) 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 lim x → 1 m √ x − 1 n √ x − 1 Here m, n ∈ N Solution. Let x = t mn ⇒ m √ x = t n and n √ x = t m x → 1 ⇒ x mn → 1 ⇒ t → 1 lim x → 1 m √ x − 1 n √ x − 1 = lim x → 1 t n − 1 t m − 1 = lim x → 1 ( t − 1)( t n − 1 + t n − 2 + · · · + t + 1) ( t − 1)( t m − 1 + t m − 2 + · · · + t + 1) = = lim x → 1 ( t n − 1 + t n − 2 + · · · + t + 1) ( t m − 1 + t m − 2 + · · · + t + 1) = n m 6 Problem S10: Give the (maximal) domain and (maximal) range of x 7 → f ( x ) := exp ( 1 cos( x ) ) and determine the left-hand and right-hand limits for all boundary points of the domain. Solution. cos( x ) 6 = 0 ⇒ x 6 = π 2 + πk, k ∈ Z D ( f ) = R \ { π 2 + πk, k ∈ Z } y = exp ( 1 cos( x ) ) ⇒ ln y = 1 cos( x ) ⇒ cos( x ) = 1 ln y − 1 ≤ 1 ln y ≤ 1 1 ln y ≤ 1 ⇒ y ∈ (0 , 1) ∪ [ e, + ∞ ) 1 ln y ≥ − 1 ⇒ y ∈ ( 0 , 1 e ] ∪ [1 , + ∞ ) So, y ∈ ( 0 , 1 e ] ∪ [ e, + ∞ ) E ( f ) = ( 0 , 1 e ] ∪ [ e, + ∞ ) lim x → π 2 − exp ( 1 cos( x ) ) = + ∞ lim x → π 2 + exp ( 1 cos( x ) ) = 0 lim x → 3 π 2 − exp ( 1 cos( x ) ) = 0 lim x → 3 π 2 + exp ( 1 cos( x ) ) = + ∞ Since f is periodic with T = 2 π : lim x → ( π 2 +2 πk ) − exp ( 1 cos( x ) ) = + ∞ lim x → ( π 2 +2 πk ) + exp ( 1 cos( x ) ) = 0 lim x → ( 3 π 2 +2 πk ) − exp ( 1 cos( x ) ) = 0 lim x → ( 3 π 2 +2 πk ) + exp ( 1 cos( x ) ) = + ∞

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