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funksional analiz misol va masalalar yechish 1 qism
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ΔB = A ∪ B, %5 A \B = A, `5 B \A = B *5 # % '5 % ` d5 # % ` \5 # ` A 1 A 2 & < ! " + #5 B 1 ∩ B 2 ⊂ (A 1 Δ B 1 ) ∪ (A 2 Δ B 2 ), %5 A 1 ΔA 2 = ∅, `5 A 1 ΔA 2 = A 1 ∪A 2 , b5 (A 1 ∪A 2 )Δ(B 1 ∪ B 2 ) ⊂ (A 1 ΔB 1 ) ∪ (A 2 ΔB 2 ) *5 # % ` '5 % ` b d5 # % ` b \5 # ` b $ f : R → R, f(x) = [0, 5 x] ' [x] x * A = [0, 8] f (A) & *5 {0; 1; 2; 3; 4; 5; 6; 7} '5 [0, 8] d5 {0; 1; 2; 3; 4; 5; 6; 7; 8} \5 {0; 1; 2; 3; 4} % f : X → [5, 26], f(x) = x 2 + 1 f − 3 " 5 X & & *5 [−5, −2] ∪ [2, 5] '5 [−2, 5] d5 [2, 5] \5 (2, 5) & f : [0, π] → [−1, 1], f(x) = cos x, g : [0, π] → [0, 1], g(x) = sin x, ϕ : [0, π 2 ] → [0, 1], ϕ(x) = sin x, ψ : [0, 2] → [0, 5], ψ(x) = x 2 + 1 ! + *5 f, g, ϕ '5 f, g, ψ d5 g, ϕ, ψ \5 f, ϕ, ψ ' f : [0, π] → [−1, 1], f(x) = cos x, g : [0, π] → [0, 1], g(x) = sin x, ϕ : [0, π 2 ] → [0, 1], ϕ(x) = sin x, ψ : [0, 2] → [0, 5], ψ(x) = x 2 + 1 ! + *5 f, g '5 f, ψ d5 f, ϕ \5 ϕ, ψ . < ! & + #5 X & %5 '! & `5 _ ! & b5 '! & *5 # ` '5 % b d5 # % ` \5 # ` b / < ! + #5 d & ! & %5 2 & & `5 = & &" *5 # ` '5 % ` d5 # \5 # % ` < ! + #5 d & ! & %5 2 & & `5 = & & *5 # ` '5 % ` d5 # \5 # % ` < ! + #5 d & ! & %5 2 & & `5 = & & *5 # ` '5 % ` d5 # \5 # % ` = A 1 A 2 & *5 A 1 = [0, 1], A 2 = (0, 1) '5 A 1 = [0, 1], A 2 = [0, 1] ∪ Q d5 A 1 = [0, ∞), A 2 = [−2, ∞) \5 A 1 = (−∞, 0], A 2 = (0, ∞) d & + & *5 Q, Z, N '5 {2; 3}, ∅ d5 {2; 3} , [0, 1], Z \5 R, Q, Z $ 2 & + & *5 Q, Z, N '5 {2; 3}, ∅ d5 {2; 3} , [0, 1], Z \5 R, Q, Z % = & + & *5 Q, Z, N '5 {2; 3}, ∅ d5 {2; 3} , [0, 1], Z \5 [0, 1], R, R\Q & [ & + & *5 Q ∼ Z ∼ N ∼ R '5 [0, 1] ∼ [0, ∞) ∼ R ∼ Q d5 [0, 1] ∼ R ∼ R 2 ∼ R\Q \5 [0, 1] ∼ R ∼ R\Q ∼ Q ' 2 ! [a, b) S : *5 _ '5 d5 \5 σ . 2 ! ! & S : *5 _ '5 d5 \5 σ / 2 ! ! & S : *5 _ '5 d5 \5 σ 2 ! ! & S : *5 _ '5 d5 \5 _ 1 ! ! S : *5 _ '5 d5 \5 σ E = {(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1} ! & S : *5 _ '5 d5 \5 σ E = [0, 1] (& ! (& & A(E) Δ && , e (e ∗ x = x ∗ e = e) & *5 e = [0, 1] '5 e = ∅ d5 e = {0} \5 e = (0, 1) $ E = [0, 1] (& ! (& & A(E) Δ && ' && x = (0, 1) & *5 x −1 = {0} '5 x −1 = ∅ d5 x −1 = (0, 1) \5 x −1 = [0, 1] % P = 1, 4 ≤ x ≤ 5 3 5 , 2, 5 ≤ y ≤ 15 2 ! ! " & *5 %> '5 21 d5 18, 75 \5 20, 4 & P = {0 ≤ x ≤ 1, 0 ≤ y ≤ 1} Q = {0, 3 ≤ x ≤ 0, 8, 0 ≤ y ≤ 1} ! ! & *5 0, 5 '5 0, 8 d5 0, 15 \5 0, 75 ' P = {0 ≤ x ≤ 1, 0 ≤ y ≤ 1} Q = {0, 5 ≤ x ≤ 1, 5, 0 ≤ y ≤ 1} ! ! & *5 1, 3 '5 1, 4 d5 1, 5 \5 2 . P = {0 ≤ x ≤ 1, 0 ≤ y ≤ 1} Q = {0, 5 ≤ x ≤ 1, 5, 0 ≤ y ≤ 1} ! ! & *5 0, 5 '5 1 d5 1, 5 \5 0, 8 / P = {0 ≤ x ≤ 5, 0 ≤ y ≤ 5} P 1 = {0 ≤ x ≤ 2, 0 ≤ y ≤ 2}. P \P 1 ! ! ! " c *5 b '5 ` d5 % \5 a A = {0 ≤ x ≤ 1, 0 ≤ y ≤ 0, 2} ∪ {0 ≤ x ≤ 1, 0, 8 ≤ y ≤ 1} & ! & *5 0, 2 '5 0, 3 d5 0, 4 \5 0, 5 A = {0 ≤ x ≤ 1, 0 ≤ y ≤ x} & ! & *5 0, 2 '5 0, 3 d5 0, 4 \5 0, 5 / ! *5 μ n ∪ k =1 A k = n k =1 μ (A k ), A i ∩ A j = ∅ i = j '5 μ ∞ ∪ n =1 A n = ∞ n =1 μ (A n ), A i ∩ A j = ∅ i = j d5 lim n →∞ μ (A n ) = μ(A), A = ∞ ∪ n =1 A n , A 1 ⊂ A 2 ⊂ · · · ⊂ A n ⊂ · · · \5 μ ∞ ∪ n =1 A n ≤ ∞ n =1 μ (A n ) / ! *5 μ n ∪ k =1 A k = n k =1 μ (A k ), A i ∩ A j = ∅ i = j '5 μ ∞ ∪ n =1 A n = ∞ n =1 μ (A n ), A i ∩ A j = ∅ i = j d5 lim n →∞ μ (A n ) = μ(A), A = ∞ ∪ n =1 A n , A 1 ⊂ A 2 ⊂ · · · ⊂ A n ⊂ · · · \5 μ ∞ ∪ n =1 A n ≤ ∞ n =1 μ (A n ) $ / ! σ − *5 μ n ∪ k =1 A k = n k =1 μ (A k ), A i ∩ A j = ∅ i = j '5 μ ∞ ∪ n =1 A n = ∞ n =1 μ (A n ), A i ∩ A j = ∅ i = j d5 lim n →∞ μ (A n ) = μ(A), A = ∞ ∪ n =1 A n , A 1 ⊂ A 2 ⊂ · · · ⊂ A n ⊂ · · · \5 μ ∞ ∪ n =1 A n ≤ ∞ n =1 μ (A n ) % / ! & *5 μ n ∪ k =1 A k = n k =1 μ (A k ), A i ∩ A j = ∅ i = j '5 μ ∞ ∪ n =1 A n = ∞ n =1 μ (A n ), A i ∩ A j = ∅ i = j d5 lim n →∞ μ (A n ) = μ(A), A = ∞ ∪ n =1 A n , A 1 ⊂ A 2 ⊂ · · · ⊂ A n ⊂ · · · \5 μ ∞ ∪ n =1 A n ≤ ∞ n =1 μ (A n ) & A ⊂ [0, 1] − ! & < ! ! & + #5 [0, 1]\A, %5 A ∩ Q, `5 A ∩ ([0, 1]\Q). *5 # ` '5 % ` d5 # % \5 # % ` ' A ⊂ [0, 1]− ! & K − = & <" ! ! & + 1) [0, 1]\A, 2) A ∩ Q, 3) A ∩ K. *5 # ` '5 % ` d5 # % \5 # % ` . μ F ! F (x) = [x] / "2 ! A = {1, 2, 3} μ F (A) *5 # '5 % d5 ` \5 > / F (x) = K(x)− = & μ F − F (x) " / "2 ! A = 1 3 ; 2 3 & ! & *5 1 3 '5 2 3 d5 π 2 \5 > # - 5 * * 0 9 / x ∈ E\ α A α , x ∈ E x / ∈ α A α ' α 0 + x A α 0 & " \ x A α 0 & ! 2 α 0 ! x ∈ E\A α 0 x ∈ α (E\A α ) ' E \ α A α ⊂ α ( Download 1.57 Mb. Do'stlaringiz bilan baham: 
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