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funksional analiz misol va masalalar yechish 1 qism
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Y × X
c #b#"#b`" X × (Y ∪ Z) = (X × Y ) ∪ (X × Z) . (X × Y ) ∪ (X 1 × Y 1 ) ⊂ (X ∪ X 1 ) × (Y ∪ Y 1 ) . (X × Y ) ∩ (X 1 × Y ) = (X ∩ X 1 ) × Y. (X × Y ) ∪ (X 1 × Y 1 ) = (X ∪ X 1 ) × (Y ∪ Y 1 ) c $ , A ∪ B = AΔ (B\A) \ ” ∪ ” ”Δ” ”\” 2 - 5 ” ∪ ” ”Δ” ” ∩ ” 4 5 ” ∩ ” ”Δ” ” ∪ ” 4 !5 ” ∩ ” ”Δ” ”\” , ” ∪ ”, ” ∩ ”, ”\”, ”Δ” - 5 ! 4 5 c % {A n } & " ! A ∗ = ∞ ∩ n =1 ∞ ∪ m =n A m , A ∗ = ∞ ∪ n =1 ∞ ∩ m =n A m , ∞ ∩ n =1 A n ⊂ A ∗ ⊂ A ∗ ⊂ ∞ ∪ n =1 A n " & * A 1 ⊂ A 2 ⊂ · · · {A n } & " ! A 1 ⊃ A 2 ⊃ · · · {A n } . ! & " ! A ∗ = A ∗ = ∞ ∪ n =1 A n , ! & " ! A ∗ = A ∗ = ∞ ∩ n =1 A n ' A n = k n : k ∈ Z − + n ∈ N ! & A ∗ = Z, A ∗ = Q . * A 3n = B, A 3n−1 = C, A 3n−2 = D, n ∈ N A ∗ A ∗ & B, C, D & $/ A kn = k − 1 n , k + 1 n , k, n ∈ N & ! ∞ ∩ n =1 ∞ ∩ k =1 A kn ; ∞ ∪ n =1 ∞ ∪ k =1 A kn ; ∞ ∪ n =1 ∞ ∩ k =1 A kn ; ∞ ∪ k =1 ∞ ∩ n =1 A kn ; ∞ ∩ k =1 ∞ ∪ n =1 A kn ; ∞ ∩ n =1 ∞ ∪ k =1 A kn & & $ , ∞ k =1 ∞ n =1 A kn ⊂ ∞ n =1 ∞ k =1 A kn {A kn } , k, n ∈ N & ! $ Ω = { x : x = (x 1 , x 2 , ... , x n , . . . )} ! " & Ω n = { x : x = (x 1 , x 2 , ..., x n , 0, 0, ...)} n + 1 − 0 " & , Ω = ∞ ∪ n =1 Ω n " $ c 0 = x : x = (x 1 , x 2 , ... ) , lim n →∞ x n = 0 − 0 ! " ! " & p ∈ N ! p = x : x = (x 1 , x 2 , ... ) , ∞ i =1 | x i | p < +∞ ( p − + ! ! " & , c 0 ⊃ ∞ ∪ p =1 p c 0 = ∞ ∪ p =1 p $ Z & Q = {η 1 , η 2 , . . . , η n , . . . } " ε > 0 ! ∞ ∪ n =1 (η n − ε, η n + ε) = R $$ ε > 0 x ∈ R m n − ε n , m n + ε n , m ∈ Z, n ∈ N \ ε> 0 m ∈Z n ∈N m n − ε n , m n + ε n = R. §. 0- 1 -0) -# - ( " ! ! (:- X & * x ∈ X f ! y = f(x) X & f ' X & f " ! E (f) & f ( E (f) = {y : y = f(x), x ∈ X}. * & & ! ( (; ' " X Y & * x ∈ X f ! Y & y X & Y & ! f ' * Y = R Y = C f X X & Y & ! f ! f : X → Y [ f : X → Y ! ! a ∈ X ! b = f(a) ∈ Y a f , X & A A & ! " Y & A & f " f (A) [ b ∈ Y X & b ! ! " b f f −1 (b) = {x ∈ X : f(x) = b} . B ⊂ Y & ! X B ! 3 ! 5 B & f f −1 (B) = {x ∈ X : f(x) ∈ B} " * ! b ∈ B ! f −1 (b) B & & , " Y & f ! & * X f : X → Y f (X) = Y + f X & Y & , ( f (X) ⊂ Y f X & Y & * f : X → Y X x 1 x 2 y 1 = f(x 1 ) y 2 = f(x 2 ) f ' f : X → Y [ f : X → Y %#"%` & f : R → R, f(x) = x 2 . ! "# f : R → R, f(x) = x 2 E (f) = [0, ∞) d ! x ∈ R ! x 2 ≥ 0 y ∈ [0, ∞) ! f (√y) = y g : R → R, g(x) = [x]. ' [x] x ! "# g : R → R, g(x) = [x] ! x ∈ R g (x) ∈ Z ( E (g) ⊂ Z ! n ∈ Z ! g (n) = n ( Z ⊂ E(g) ' E (g) = Z \ D : R → R, D(x) = ⎧ ⎨ ⎩ 1, agar x ∈ Q 0, agar x ∈ R\Q. (2.1) ! "# \ D : R → R " E (D) = {0, 1} & %#" f A = [0, 3) & 3 5 B = (1, 4) & & ! "# f (x) = x 2 R + = [0, ∞) ! " ! f ([0, 3)) = [0, 9) [ B = (1, 4) & f & x ∈ R : x 2 ∈ (1, 4) 1 < x 2 < 4 ! & f −1 (B) ' ! (−2, −1) ∪ (1, 2) & \ f −1 (B) = (−2, −1) ∪ (1, 2) $ %`" D A = R\Q & B = (1, ∞) & & ! "# D R\Q & ! ! D(R\Q) = {0}. \ # + \ D −1 (B) = ∅. % f : R → R, f(x) = ax + b, a = 0 d f : R → R ! c ∈ R ax +b = c ! " _ ! + f : R → R, ! ( ' ! x = c − b a & * f : X → Y A ⊂ X ! f : A → B (B = f(A)) f (A) = B ! f : X → Y ! ' & ( f (A ∪ B) = f(A) ∪ f(B). (2.2) * y ∈ f(A ∪ B) y = f(x) x A B & 2 " y ∈ f(A) ∪ f(B). ' f (A ∪ B) ⊂ f(A) ∪ f(B). [ y ∈ f(A) ∪ f(B) , y = f(x) x A B & ( Download 1.57 Mb. Do'stlaringiz bilan baham: 
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