f:X→Y
•f
•A, B X f(A∩B) = f(A)∩f(B)
1 =⇒2y=f(A∩B)x∈A∩B
y=f(x)y∈f(A)y=f(x)x∈A y ∈f(B)
y∈f(A)∩f(B)y∈f(A)∩f(B)a∈A y =f(a)b∈B
y=f(b)f a =b a ∈A∩B y ∈f(A∩B)
2 =⇒1a b f(a) = f(b) = y A ={a}B={b}f(A) = f(B) = {y}
f(A∩B) = {y}A∩B6=∅a=b
f:E→F f A P(E)f(A) = f(A)
A A
f A P(E)
x f(A)x=f(y)y∈A x ∈f(A)x=f(z)
z∈A f(y) = f(z)f y =z y A z
x /∈f(A)f(A)⊂f(A)
x∈f(A)f y E
x=f(y)y A x f(A)y
A x ∈f(A)
AP(E)f(A) = f(A)
f x, y f(x) = f(y)x6=y
A={x}y∈A f(y)∈f(A)⊂f(A)f(A) = {f(x)}f(y)6=f(x)
f A =E f(E) = f(E)f(E) = f(∅) = ∅
f(E) = ∅f(E) = F f
EP(E)A B E
f:P(E)→ P(A)× P(B)
X7→ (X∩A, X ∩B).
f A ∪B=E
f A ∩B=∅
A B f
A∪B6=E x ∈E\(A∪B)X={x}
f(X)=(X∩A, X ∩B)=(∅,∅)x A B f(∅)=(∅,∅)
f(X) = f(∅)X6=∅f
X⊂E A ∪B=E
X=X∩E=X∩(A∪B) = (X∩A)∪(X∩B).
X, X0⊂E f(X) = f(X0)X∩A=X∩A0X∩B=X∩B0
X= (X∩A)∪(X∩B) = (X0∩A)∪(X0∩B) = X0.
f
f x ∈A X ⊂E f(X) = ({x},∅)
X∩B=∅x∈X∩A x ∈X x /∈B A ∩B=∅
A∩B=∅A0⊂A B0⊂B X =A0∪B0A∩B=∅
X∩A=A0X∩B=B0f(X) = (A0, B0)f
f A∪B=E A ∩B=∅(A, B)
E(A0, B0)7→ A0∪B0
f:N2→N∗(n, p)7→ 2n(2p+ 1) fN2
N