Seminari 1 Conjunts

8/17/2019 Seminari 1 Conjunts

1/16

E

P (E )

E

∅

X

Y

E

X ∩ Y = X ⇐⇒ X ⊆ Y

X ∩ Y = X =⇒ X ⊆ Y

X ∩ Y = X

x ∈ X X ∩ Y = X x ∈ X ∩ Y x ∈ X x ∈ Y

x ∈ Y

x ∈ X

∀x ∈ X, x ∈ Y

X ⊆ Y

X ∩ Y = X ⇐= X ⊆ Y

X ⊆ Y

X ∩ Y = X

x ∈ X ∩ Y x ∈ X x ∈ Y x ∈ X

∀x ∈ X ∩ Y, x ∈ X X ∩ Y ⊆ X

x ∈ X X ⊆ Y x ∈ Y x ∈ X x ∈ Y

x ∈ X ∩ Y ∀x ∈ X, x ∈ X ∩ Y X ⊆ X ∩ Y

X ∩ Y = X

X ∪ Y = X ⇐⇒ Y ⊆ X

X ∪ Y = X =⇒ Y ⊆ X

X ∪ Y = X

x ∈ Y x ∈ X ∪ Y X ∪ Y = X x ∈ X ∀x ∈ Y, x ∈ X Y ⊆ X

X ∪ Y = X ⇐= Y ⊆ X

Y ⊆ X

X ∪ Y = X

x ∈ X ∪ Y x ∈ X x ∈ Y Y ⊆ X x ∈ Y ⇒ x ∈ X

x ∈ X x ∈ Y ⇒ x ∈ X x ∈ X x ∈ X ∪ X = X ∀x ∈ X ∪ Y, x ∈ X Y ⊆ X

x ∈ X x ∈ X x ∈ Y x ∈ X ∪ Y ∀x ∈ Y, x ∈ X ∪ Y X ⊆ X ∪ Y

X ∪ Y = X


2/16

f : P (E ) → P (E ) f (X ) = (X ∩ A) ∪ B A ⊆ E B ⊆ E

E

A = ∅ f (X ) X ∈ P (E )

f (X ) Def.

= (X ∪ A) ∩ B A=∅

= (X ∪ ∅) ∩ B (1.b)

= ∅ ∩ B (1.a)

= B

B = E

f (X ) Def.

= ((X ∩A)∪B B=E

= ((X ∩A)∪E Distributiva

= (X ∪E )∩(A∪E ) (1.b)

= E ∩E (1.a)

= E

f (∅) f (A) f (B) f (E )

f (∅) = (∅ ∩ A) ∪ B = ∅ ∪ B = B

f (A) = (A ∩ A) ∪ B = A ∪ B

f (B) = (B ∩ A) ∪ B = B (B ∩ A) ⊆ Bf (E ) = (E ∩ A) ∪ B = A ∪ B

f

∀X, X ∈ P (E ), X ⊆ X =⇒ f (X ) ⊆ f (X )

X ⊆ X

x ∈ f (X ) x ∈ (X ∩ A) ∪ B x ∈ (X ∩ A) ∪ B ⇔ x ∈ (X ∩ A)

x ∈ B ⇔ (x ∈ X x ∈ A) x ∈ B X ⊆ X x ∈ X ⇒ x ∈ X (x ∈ X

x ∈ A) x ∈ B ⇒ (x ∈ X

x ∈ A)

x ∈ B ⇔ x ∈ (X ∩ A) ∪ B

x ∈ f (X ) ∀x ∈ f (X ), x ∈ f (X )

f (X ) ⊆ f (X )

Y ∈ P (E )

Y

f

P (E )

B ⊆ Y ⊆ A ∪ B

f (Y ) = Y

⇒ ⇒ ⇒


3/16

⇒ Y f P (E ) ∃X ∈ P (E ) Y = f (X )

x ∈ Y ∃X ∈ P (E ) Y = f (X ) f (X ) = (X ∩ A) ∪ B =

(X ∪ B) ∩ (A ∪ B) x ∈ (X ∪ B) ∩ (A ∪ B) x ∈ X ∪ B x ∈ A ∪ B

x ∈ A ∪ B ∀x ∈ Y , x ∈ A ∪ B Y ⊆ A ∪ B

x ∈ B ∀X ∈ P (E ), x ∈ (X ∩ A) ∪ B = f (X ) ∃X ∈ P (E ) Y = f (X )

x ∈ Y ∀x ∈ B , x ∈ Y

B ⊆ Y

∃X ∈ P (E )

Y = f (X ) ⇒ B ⊆ Y ⊆ A ∪ B

⇒ B ⊆ Y ⊆ A ∪ B

f (Y ) = Y

x ∈ f (Y ) f x ∈ (Y ∩ A) ∪ B x ∈ Y ∩ A x ∈ B

x ∈ Y ∩ A x ∈ Y x ∈ A x ∈ Y

x ∈ B B ⊆ Y x ∈ Y

∀x ∈ f (Y ), x ∈ Y f (Y ) ⊆ Y

x ∈ Y B ⊆ Y Y = Y ∪ B Y = Y ∪ B ⊆ A ∪ B Y = (Y ∪ B) ∩ (A ∪ B) = (Y ∩ A) ∪ B = f (Y ) x ∈ f (Y ) ∀x ∈ Y, x ∈ f (Y ) Y ⊆ f (Y )

B ⊆ Y ⊆ A ∪ B ⇒ f (Y ) = Y

⇒ Y = f (Y )

Y ∈ P (E ) ∃X ∈ P (E ) Y = f (X ) X = Y

Y = f (Y ) ⇒ ∃X ∈ P (E ) Y = f (X )

f of = f

f of (X ) = f (f (X )) = f ((X ∩ A) ∪ B) = f ((X ∪ B) ∩ (A ∪ B)) = (((X ∪ B) ∩ (A ∪ B)) ∩

A)∪B = ((X ∪B)∩((A∪B)∩A))∪B (1.a)

= ((X ∪B)∩A)∪B = ((X ∪B)∪B)∩(A∪B) (1.b)

=(X ∪ B) ∩ (A ∪ B) = (X ∩ A) ∪ B = f (X )


4/16

P (E ) X : f (X ) = A

B A

∃x ∈ B x ∈ A ∃x ∈ B x ∈ A ⇒ ∃x ∈ (X ∩ A) ∪ B x ∈ A ⇔ ∃x ∈ f (X ) x ∈ A ∃x ∈ f (X ) x ∈ A

f (X )

A ∀X ∈ P (E ), f (X ) = A

B ⊆ A f (X ) = (X ∩ A) ∪ B = (X ∪ B) ∩ (A ∪ B)

(1.b)= (X ∪ B) ∩ A

f (X ) = A ⇔ (X ∪ B) ∩ A = A (1.a)

⇔ A ⊆ X ∪ B

B ⊆ A

A ⊆ X ∪ B ⇔ A B ⊆ X

A ⊆ X ∪ B

x ∈ A B x ∈ A x ∈ B A ⊆ X ∪ B x ∈ X ∪ B x ∈ B ⇔ x ∈ X ∪ B x ∈ Bc ⇔ x ∈ (X ∪ B) ∩ Bc = (X ∩ Bc) ∪ (B ∩ Bc) =(X ∩ Bc) ∪ ∅ = X ∩ Bc x ∈ X x ∈ Bc x ∈ X ∀x ∈ A B, x ∈ X A B ⊆ X

A B ⊆ X

x ∈ A x ∈ A ⇒ x ∈ A∪B = (A∪B)∩E = (A∪B)∩(Bc∪B) =

(A ∩ Bc) ∪ B = (A B) ∪ B x ∈ A B x ∈ B A B ⊆ X x ∈ X x ∈ B x ∈ X ∪ B ∀x ∈ A, x ∈ X ∪ B A ⊆ X ∪ B

A ⊆ X ∪ B ⇔ A B ⊆ X

X : f (X ) = A

• B A • X : A B ⊆ X B ⊆ A

P (E ) X : f (X ) = B

f (X ) = B ⇔ (X ∩ A) ∪ B = B ⇔ X ∩ A ⊆ B X : f (X ) = B

X : X ∩ A ⊆ B

X ∩ A ⊆ B ⇔ X ⊆ B ∪ Ac

X ∩ A ⊆ B

x ∈ X = X ∩ E = X ∩ (A ∪ Ac) = (X ∩ A) ∪ (X ∩ Ac)

x ∈ X ∩ A

x ∈ X ∩ Ac

• x ∈ X ∩ A X ∩ A ⊆ B x ∈ X ∩ A ⇒ x ∈ B ⇒ x ∈ B x ∈ Ac x ∈ B ∪ Ac

• x ∈ X ∩ Ac x ∈ X ∩ Ac ⇒ x ∈ X x ∈ Ac ⇒ x ∈ Ac ⇒ x ∈ B x ∈ Ac ⇔ x ∈ B ∪ Ac x ∈ B ∪ Ac


5/16

∀x ∈ X, x ∈ B ∪ Ac X ⊆ B ∪ Ac

X ⊆ B ∪ Ac

x ∈ X ∩ A x ∈ X ∩ A ⇔ x ∈ X x ∈ A ⇒ x ∈ B ∪ Ac x ∈ A ⇔ x ∈

(B ∪ Ac) ∩ A = (B ∩ A) ∪ (Ac ∩ A) = (B ∩ A) ∪ ∅ = B ∩ A ⇔ x ∈ B x ∈ A ⇒ x ∈ B

x ∈ B

∀x ∈ X ∩ A, x ∈ B X ∩ A ⊆ B

X ∩ A ⊆ B ⇔ X ⊆ B ∪ Ac

X : B ∩ Ac ⊆ X

A

B

f

f ∃C ∈ P (E ), ∀X ∈ P (E ), f (X ) = C

f

⇔ ∀X ∈ P (E ), f (X ) = B

f ∃C ∈ P (E ), ∀X ∈ P (E ), f (X ) = C

X = ∅ f (∅) = (∅ ∩ A) ∪ B = ∅ ∪ B = B C = B ∀X, f (X ) = B

∀X ∈ P (E ), f (X ) = B B ∈ P (E ) f ∃C ∈ P (E ), ∀X ∈ P (E ), f (X ) = C C = B

∀X ∈ P (E ), f (X ) = B ⇔ A ⊆ B

∀X ∈ P (E ), f (X ) = B

x ∈ A x ∈ A ⇒ x ∈ A x ∈ B ⇔ x ∈ A ∪ B = (A ∩ A) ∪ B =

f (A) = B ∀x ∈ A, x ∈ B A ⊆ B

A ⊆ B

X ∈ P (E ) f (X ) = (X ∩A)∪B = (X ∪B)∩(A∪B)

(1.b)= (X ∪B)∩B

(1.a)=

B ∀X ∈ P (E ), f (X ) = B

f

A ⊆ B

A B f

f

X, Y ∈ P (E ), X = Y ⇔ f (X ) = f (Y )

f (X ) = f (Y ) ⇔ X = Y

f ⇔ A = E B = ∅

f

f (X ) = f (Y ) ⇔ X = Y

f (E ) = (E ∩ A) ∪ B = A ∪ B = (A ∩ A) ∪ B = f (A)

f A = E


6/16

f (∅) = (∅ ∩ A) ∪ B = ∅ ∪ B = B = B ∪ B = (B ∩ E ) ∪ B =

(B ∩ A) ∪ B = f (B) f B = ∅

A = E

B = ∅

∀X ∈ P (X ), f (X ) = (X ∩ A) ∪ B = (X ∩ E ) ∪ ∅ = X ∪ ∅ = X f (X ) = X f f (X ) = f (Y ) ⇔ X = Y

X = Y ⇔ X = Y

A B f

f

∀Y ∈ P (E ), ∃X ∈ P (E ), f (X ) = Y

f ⇔ A = E B = ∅

f ∀Y ∈ P (E ), ∃X ∈ P (E ), f (X ) = Y

Y = ∅ ∃X ∈ P (E ), f (X ) = (X ∩ A) ∪ B = ∅

B = ∅ B (X ∩A)∪B

(X ∩ A) ∪ B = ∅

Y = E

∃X ∈ P (E ), f (X ) = (X ∩ A) ∪ B =(X ∩ A) ∪ ∅ = X ∩ A = E A = E A ⊂ E

X ∩ A ⊂ E

X ∩ A = E

A = E

B = ∅

∀X ∈ P (X ), f (X ) = (X ∩ A) ∪ B = (X ∩ E ) ∪ ∅ = X ∪ ∅ = X

f (X ) = X f ∀Y ∈ P (E ), ∃X ∈ P (E ), f (X ) =Y

X = Y

f f


7/16

E A B E

A ∗ B = (A ∩ B) ∪ (Ac ∩ Bc),

Ac A E P (E )

A B E A ∗ B

A ∗ B = B ∗ A

A ∗ B

E

A ∗ B = (A ∩ B) ∪ (Ac ∩ Bc) = (B ∩ A) ∪ (Bc ∩ Ac) = B ∗ A

A B C E

A ∗ (B ∗ C ) = (A ∗ B) ∗ C

A ∗ (B ∗ C )

A B C B ∩ C Bc ∩ C c B ∗ C A ∩ (B ∗ C ) Ac ∩ (B ∗ C )c A ∗ (B ∗ C )

(A ∗ B) ∗ C


8/16

A B C A ∩ B Ac ∩ Bc A ∗ B (A ∗ B) ∩ C (A ∗ B)c ∩ C c (A ∗ B) ∗ C

A ∗ (B ∗ C ) (A ∗ B) ∗ C

A E A ∗ E A ∗ A

∗

• A ∗ E = (A ∩ E ) ∪ (Ac ∩ E c) = A ∪ (Ac ∩ ∅) = A ∪ ∅ = A

• A ∗ A = (A ∩ A) ∪ (Ac ∩ Ac) = A ∪ Ac = E

A B C E

A ∗ B = Ac ∗ Bc

Ac∗Bc = (Ac∩Bc)∪((Ac)c∩(Bc)c) = (Ac∩Bc)∪(A∩B) = (A∩B)∪(Ac∩Bc) = A∗B

(A ∗ B)c = A ∗ (Bc) = (Ac) ∗ B

(A ∗ B)c = (Ac ∩ B) ∪ (A ∩ Bc)

(A ∗ B)c = ((A ∩ B) ∪ (Ac ∩ Bc))c = (A ∩ B)c ∩ (Ac ∩ Bc)c = (Ac ∪ Bc) ∩ ((Ac)c ∪(Bc)c) = (Ac ∪ B c) ∩ (A ∪ B ) = (Ac ∩ A) ∪ (Ac ∩ B ) ∪ (Bc ∩ A) ∪ (Bc ∩ B ) =∅ ∪ (Ac ∩ B) ∪ (Bc ∩ A) ∪ ∅ = (Ac ∩ B) ∪ (Bc ∩ A) = (Ac ∩ B) ∪ (A ∩ Bc)

(Ac ∪ B) ∩ (A ∪ Bc) = A ∗ (Bc)

(Ac

∩ B) ∪ (A ∩ Bc

) = (A ∩ Bc

) ∪ (Ac

∩ B) = (A ∩ Bc

) ∪ (Ac

∩ (Bc

)c

) = A ∗ (Bc

)

(Ac ∪ B) ∩ (A ∪ Bc) = (Ac) ∗ B

(Ac ∪ B) ∩ (A ∪ Bc) = (Ac ∪ B) ∩ ((Ac)c ∪ Bc) = (Ac) ∗ B

(A ∗ B)c = A ∗ (Bc) = (Ac) ∗ B

E

A ∗ X = B

X = A ∗ B A ∗ X = B ⇔ A ∗ B = X

A ∗ X = B ⇒ A ∗ B = X


9/16

A ∗ X = B

A ∗ B ⊆ X

A ∗ X = B X ∪ (A ∗ B)c = E

X ∪ (A ∗ B)c = X ∪ ((Ac ∪ Bc) ∩

(A ∪ B)) = (X ∪ (Ac ∪ Bc)) ∩ (X ∪ (A ∪ B)) = ((X ∪ Ac) ∪ Bc) ∩ ((X ∪ A) ∪ B)

(X ∪ Ac) ∪ Bc (X ∪ A) ∪ B

(X ∪ Ac) ∪ Bc

X A∗X = (A∩X )∪(Ac∩X c) = (A∪X c)∩(X ∪Ac) = B

x ∈ B ⇔ x ∈ (A ∪ X c) x ∈ (X ∪ Ac) ⇒ x ∈ (X ∪ Ac)

B ⊆ X ∪ Ac

(X ∪ Ac) ∪ Bc ⊇ B ∪ Bc = E E X,A,B ⊆ E

E

(X ∪ Ac) ∪ Bc ⊆ E

(X ∪ Ac) ∪ Bc = E

(X ∪ A) ∪ B

X

A ∗ X = B A ∗ X = B ⇔ (A ∗ X )c = (Ac ∪

X c)∩(A∪X ) = Bc x ∈ Bc ⇔ x ∈ Ac∪X c x ∈ A ∪X ⇒ x ∈ A ∪X = X ∪A Bc ⊆ X ∪ A

(X ∪ A) ∪ B ⊇ Bc ∪ B = E E X,A,B ⊆ E

E (X ∪ A) ∪ B ⊆ E

(X ∪ A) ∪ B = E

((X ∪ Ac) ∪ Bc) ∩ ((X ∪ A) ∪ B) =E ∩ E = E X ∪ (A ∗ B)c = E

X ∪ (A ∗ B)c = E ⇒A ∗ B ⊆ X

X ∪ (A ∗ B)c = E

x ∈ A∗B A∗B = E ∩(A∗B) = (X ∪(A∗B)c)∩(A∗B) = X ∩(A∗B) x ∈ X ∩ (A ∗ B) x ∈ X x ∈ A ∗ B x ∈ X

A ∗ X = B ⇒ X ⊆ A ∗ B

X ∩ (A ∗ B)c = ∅

X ∩ (A ∗ B)c = X ∩ ((A ∩ Bc) ∪

(Ac ∩ B)) = (X ∩ (A ∩ Bc)) ∪ (X ∩ (Ac ∩ B)) = ((X ∩ A) ∩ Bc) ∪ ((X ∩ Ac) ∩ B)

(X ∩ A) ∩ Bc (X ∩ Ac) ∩ B

(X ∩ A) ∩ Bc

x ∈ X ∩ A x ∈ X ∩ A ⇒ x ∈ A ∩ X x ∈ Ac ∩ X c ⇔ x ∈(A ∩ X ) ∪ (Ac ∩ X c) = A ∗ X = B X ∩ A ⊆ B


10/16

(X ∩ A) ∩ Bc ⊆ B ∩ Bc = ∅

∅ ⊆ (X ∩A)∩Bc (X ∩A)∩Bc =∅

(X ∩ Ac) ∩ B

x ∈ X ∩ Ac x ∈ X ∩ Ac ⇒ x ∈ (X ∩ Ac) (X c ∩ A) ⇔ x ∈

(X ∩Ac)∪(X c∩A) (2.b)

= (X ∗A)c = (A∗X )c = Bc X ∩Ac ⊆ Bc

(X ∩ Ac) ∩ B ⊆ Bc ∩ B = ∅

∅ ⊆ (X ∩Ac)∩B (X ∩Ac)∩B =∅

((X ∩ A) ∩ Bc) ∪ ((X ∩ Ac) ∩ B) =

∅ ∪ ∅ = ∅ X ∩ (A ∗ B)c = ∅

X ∩ (A ∗ B)c = ∅ ⇒X ⊆ A ∗ B

X ∩ (A ∗ B)c = ∅

x ∈ X = X ∩ E = X ∩ ((A ∗ B) ∪ (A ∗ B)c) = (X ∩ (A ∗ B)) ∪ (X ∩ (A ∗ B)c) =(X ∩ (A ∗ B)) ∪ ∅ = (X ∩ (A ∗ B)) x ∈ X x ∈ A ∗ B x ∈ A ∗ B X ⊆ A ∗ B

A∗B ⊆ X X ⊆ A ∗B

X = A ∗ B

A ∗ X = B ⇐ A ∗ B = X

X

A ∗ X = A ∗ (A ∗ B) (1.b)

= (A ∗ A) ∗ B (1.c)

= E ∗ B = (E ∩ B) ∪ (E c ∩ Bc) = B ∪ (∅ ∩ Bc) =B ∪ ∅ = B

A ∗ X = B ⇔ A ∗ B = X

A ∗ X = B X = A ∗ B


11/16

E F

E

F

F

E

E

F

f : E → F g : F → E

h = g ◦ f, R = E \ g(F ), F = M ∈ P (E ) | R ∪ h(M ) ⊆ M , A =M ∈F

M.

N = N ∪ {0}

X

Y

u : X −→ Y U, V ⊆ X

U ⊆ V ⇒ u(U ) ⊆ u(V ) u(U ∪ V ) = u(U ) ∪ u(V )

u

u(U ∩ V ) = u(U ) ∩ u(V )

E ∈ F A ∈ F

E ∈ F R ∪ h(E ) ⊆ E

x ∈ R ∪ h(E ) = E \ g(f ) ∪ h(E )

x ∈ E \ g(f )

x ∈ h(E ) ⇒ x ∈ E

x ∈ E ⇔ x ∈ E x ∈ E R ∪ h(E ) ⊆ E

F E ∈ F ⇔ R ∪ h(E ) ⊆ E E ∈ F

A ∈ F R ∪ h(A) ⊆ A

x ∈ R ∪ h(A) = R ∪ h(

M ∈F M ) h

x ∈ R ∪

M ∈F h(M )

x ∈ M ∈F (R ∪ h(M )) M ∈ F ⇔ R ∪ h(M ) ⊆M

x ∈

M ∈F (R ∪ h(M )) ⇒ x ∈

M ∈F M = A

x ∈ A

R ∪ h(A) ⊆ A

A ∈ F ⇔ R ∪ h(A) ⊆ A

A ∈ F

∀M ∈ F , R ∪ h(M ) ∈ F

∀M ∈ F , R∪h(M ) ∈ F ∀M ∈ F , R∪h(R∪h(M )) ⊆R ∪ h(M ) F

M ∈ F R ∪ h(M ) ⊆ M


12/16

x ∈ R ∪ h(R ∪ h(M )) x ∈ R x ∈ h(R ∪ h(M )) R ∪ h(M ) ⊆ M

x ∈ R x ∈ h(M ) x ∈ R ∪ h(M ) R ∪ h(R ∪ h(M )) ⊆ R ∪ h(M ) ∀M ∈ F , R ∪ h(R ∪ h(M )) ⊆ R ∪ h(M ) ∀M ∈ F , R ∪ h(M ) ∈ F

A =

n≥0 h

n(R)

A ⊇

n≥0 hn(R)

∀n ∈ N, A = (n

i=0 hi(R)) ∪ hn+1(A)

n = 0 (n

i=0 hi(R)) ∪ hn+1(A) = R ∪ h(A) = A

A ∈ F R ∪ h(A) ⊆ A A ∈ F R ∪ h(A) ∈ F A =

M ∈F M =

M ∈F M ∩ (R ∪ h(A)) = A ∩ (R ∪ h(A))

A = A ∩ (R ∪ h(A)) ⇔ A ⊆ R ∪ h(A) A ⊆ R ∪ h(A)

n A = (n

i=0 hi(R)) ∪ hn+1(A)

n + 1 A = (n+1

i=0 hi(R)) ∪ h(n+1)+1(A)

A = (n

i=0 hi(R)) ∪ hn+1(A) = (

ni=0 h

i(R)) ∪ hn+1(R ∪ h(A)) = (n

i=0 hi(R)) ∪ hn+1(R) ∪

hn+1(h(A)) = (n+1

i=0 hi(R)) ∪ h(n+1)+1(A))

A ⊇ n≥0 hn(R)

x ∈

n≥0 h

n(R)

x ∈

n≥0 hn(R) ⇒ ∃ j ∈ N, x ∈ h j(R) ⇒ x ∈ (

ji=0 h

i(R)) ∪

h j+1 ∀n ∈ N, A = (

ni=0 h

i(R))

A = ( j

i=0 hi(R))∪h j+1 x ∈ A

A ⊇

n≥0 hn(R)

A ⊆

n≥0 h

n(R)

n≥0 h

n(R) ∈ F n≥0 h

n(R) = R ∪

n≥1 hn(R) = R ∪

n≥0 h(h

n(R)) = R ∪ h(

n≥0 hn(R))

R ∪ h(

n≥0 h

n(R)) ⊆

n≥0 h

n(R)

n≥0 h

n(R) ∈ F ⇔ R ∪ h(

n≥0 h

n(R)) ⊆

n≥0 hn(R) n≥0 hn(R) ∈ F

n≥0 h

n(R) ∈ F

A =

M ∈F M =

M ∈F M ∩

n≥0 hn(R) =

A ∩

n≥0 hn(R)

A = A ∩

n≥0 h

n(R) ⇔ A ⊆

n≥0 hn(R)

A ⊆

n≥0 hn(R)

A ⊇

n≥0 h

n(R) A ⊆

n≥0 h

n(R)

A =

n≥0 hn(R)


13/16

B A E A = f (A) B = g−1(B) R ∪ h(A) = A B A F

∀n ∈ N, A = (n

i=0 hi(R)) ∪ hn+1(A)

n = 0

A = R ∪ h(A)

(A)c = B

(A)c = (f (A))c = (f (

n≥0 h

n(R)))c = (

n≥0 f (hn(R)))c =

n≥0(f (hn(R)))c

B = g−1(B) = g−1(Ac) = g−1((R ∪ h(A))c) = g−1(Rc ∩ h(A)c) =g−1(Rc) ∩ g−1(h(A)c)

Rc = (E \ g(F ))c = (E ∩ (g(F ))c)c = E c ∪ ((g(F ))c)c = ∅ ∪ g(F ) = g(F ) g−1(Rc) = g−1(g(F ))

{x ∈ F | g(x) ∈ g(F )} g ∀x ∈ F, ∃! y ∈ E g(x) = y y ∈ {g(x) ∈ E | x ∈ F } = g(F ) ∀x ∈ F, ∃! y ∈ g(F ) g(x) = y {x ∈ F | g(x) ∈ g(F )} = {x ∈ F } = F g−1(Rc) = F

g−1(Rc)∩g−1(h(A)c) = F ∩g−1(h(A)c) = g−1(h(A)c) = g−1((h(

n≥0 hn(R)))c) =

g−1((

n≥0 h(hn(R)))c) = g−1(

n≥0 h

n+1(R)c) =

n≥0 g−1(hn+1(R)c)

B =

n≥0 g−1(hn+1(R)c)

(f (A))c = B ⇔ n≥0(f (h

n(R)))c =n≥0 g

−1

(hn+1

(R)c

)

n≥0(f (h

n(R)))c =

n≥0 g−1(hn+1(R)c)

∀n ∈ N, (f (hn(R)))c = g−1(hn+1(R)c)

n ∈ N a ∈ g−1(hn+1(R)c)

a ∈ {x | g(x) ∈ hn+1(R)c} = {x | g(x) ∈ hn+1(R)} = {x | g(x) ∈ h(hn(R))} = {x | g(x) ∈g(f (hn(R)))} a ∈ {x | g(x) ∈ g(f (hn(R)))} ⇔ g(a) ∈ g(f (hn(R))) = {g(x) | x ∈f (hn(R))} ⇔ x ∈ f (hn(R)) g(x) = g(a) g x ∈ f (hn(R))

g(x) = g(a) ⇔ x ∈ f (hn(R)) x = a ⇔ a ∈ f (hn(R)) ⇔ a ∈ (f (hn(R)))c a ∈ (f (hn(R)))c

∀n ∈ N, (f (hn(R)))c = g−1(hn+1(R)c)

(f (A))c = B

f : A → A g : B → B f g

f g


14/16

f x, y ∈ A f (x) = f (y) ⇔

x = y

x, y ∈ A x, y ∈ E f f f (x) = f (x)

f (y) = f (y) f f (x) = f (y) ⇔ x = y f (x)

f (x) f (y)

f (y)

f (x) = f (y) ⇔ x = y

f

∀y ∈ A, ∃x ∈ A f (x) = y

y ∈ A y ∈ f (A) y ∈ {f (x) | x ∈ A}

y ∈ {f (x) | x ∈ A} ⇔ ∃x ∈ A f (x) = y ∃x ∈ A f (x) = y

y ∈ A

∀y ∈ A, ∃x ∈ A f (x) = y

g

x, y ∈ B g(x) = g (y) ⇔ x = y

x, y ∈ B x, y ∈ F g g g(x) = g(x)

g(y) = g(y) g g(x) = g(y) ⇔ x = y g(x)

g(x) g(y)

g(y)

g(x) = g (y) ⇔ x = y

g ∀y ∈ B, ∃x ∈ B g(x) = y

y ∈ B y ∈ Ac = (R∪h(A))c = Rc∩h(A)c Rc = g(F )

y ∈ g(F )∩h(A)c y ∈ g(F ) y ∈ h(A)c y ∈g(F ) y ∈ g(F ) = {g(x) | x ∈ F } y ∈ {g(x) | x ∈ F } ⇔ ∃x ∈ F

g(x) = y

∃x ∈ F

g(x) = y

g(x) = y ⇒ ∃a ∈ B

g(x) = a

a = y ⇔ x ∈ {x | g(x) ∈ B } = g−1(B) = B

x ∈ B ∃x ∈ B g(x) = y x ∈ B g(x) = g (x) ∃x ∈ B g(x) = y y ∈ B ∀x ∈ B, ∃x ∈ B g(x) = y

f

g

φ : E −→ F

x −→

f (x) x ∈ A(g)−1(x)

x ∈ B

g

(g)−1

φ ∀x ∈ E , ∃! y ∈ F φ(x) = y

x ∈ E = A ∪ Ac x ∈ A x ∈ Ac


15/16

x ∈ A φ(x) = f (x) f ∃! y ∈ A f (x) = y A ⊆ F ∃! y ∈ F φ(x) = y

x ∈ Ac φ(x) = (g)−1(x) (g)−1 ∃! y ∈ B

(g)−1(x) = y B ⊆ F ∃! y ∈ F φ(x) = y

A ∩ Ac = ∅ x

∃! y ∈ F φ(x) = y

x ∀x ∈ E , ∃! y ∈ F φ(x) = y φ

φ

φ

φ

x, y ∈ E , φ(x) = φ(y) ⇔ x = y

x, y ∈ E = A ∪ Ac = A ∪ B

• x ∈ A y ∈ A φ(x) = f (x) φ(y) = f (y) φ(x) = φ(y) ⇔ f (x) =f (y) f f (x) = f (y) ⇔ x = y φ(x) = φ(y) ⇔ x = y

• x ∈ B y ∈ B φ(x) = (g)−1(x) φ(y) = (g)−1(y) φ(x) = φ(y) ⇔

(g

)

−1

(x

) = (g

)

−1

(y

)

(g

)

−1

(g

)

−1

(x

) = (g

)

−1

(y

) ⇔x = y

φ(x) = φ(y) ⇔ x = y

• x ∈ A y ∈ B A ∩ B = A ∩ Ac = ∅ x = y φ(x) = f (x) φ(y) = (g)−1(y) f (g)−1

φ(x) ∈ A φ(y) ∈ B A ∩ B = A ∩ (A)c = ∅

φ(x) = φ(y) x = y φ(x) = φ(y)

x = y ⇔ φ(x) = φ(y)

φ(x) = φ(y) ⇔ x = y

• x ∈ B y ∈ A x ∈ A y ∈ B

x

y y

x

x, y ∈ E , φ(x) = φ(y) ⇔ x = y φ

φ ∀y ∈ F, ∃x ∈ E φ(x) = y

y ∈ F y ∈ F = A ∪ (A)c y ∈ A y ∈ (A)c = B


16/16

• y ∈ A f

y ∈ A ∃x ∈ A f (x) = y x ∈ A φ(x) = f (x) ∃x ∈ A φ(x) = y A ⊆ E ∃x ∈ E φ(x) = y

• x ∈ B (g)−1 y ∈ B ∃x ∈ B (g)−1(x) = y x ∈ B φ(x) = (g)−1(x) ∃x ∈ B φ(x) = y B ⊆ E ∃x ∈ E φ(x) = y

∃x ∈ E φ(x) = y y ∀y ∈ F, ∃x ∈ E φ(x) = y φ

φ

E F

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