U1_Conjuntos_relaciones_y_funciones

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Si un objeto elemento e pertenece a un objeto conjunto C lo denotaremos e ∈ C y se lee: e pertenece a C. 
A = {x ∈ N x⁄ = 2n con n ∈ N} 
• 
• 
A = {2,4,6,8,10,...} 
{x ∈ A x⁄ ≠ 
x,con A conjunto} 
A = B 
2N 
2N = {2,4,6,8,10,...} 
A 
2N 
2N 
P(∅) = {∅} 
P(P(∅)) = {{∅},∅} 
A ⊂ B y B ⊂ A 
A = {a,b,c} 
A = {a,b,c} 
P(A) = 
P(A) = 
P(A) = 
P(A) = {X X⁄ ⊂ A} 
A ⊂ B o B ⊃ A 
U 
∅ o { } 
{A,{a},{b},{c},{b,c},{a,b},{c,a},∅} 
Teoría de conjuntos 
P(A) 
• 
• 
• 
• 
A ∩ 
B El símbolo se usará como una relación entre conjuntos,el de la izquierda 
y el de la derecha 
{xεU ∕ x ∈ A ∨ x ∈ B} 
{x ∈ U x⁄ ∈ A ∧ x ∈ B} 
{x ∈ U x⁄ ∈ A ∧ x ∈ B} 
A 
{x ∈ U x⁄ ∉ A} 
{x ∈ U x⁄ ∉ A} 
A ∪ B 
A ∪ B 
A o A ∩ A = A 
∪ A = A 
o A ∩ B = B ∩ A, A ∪ B = B ∪ A 
o A ∩ (B ∩ C) = (A ∩ B) ∩ C, A ∪ (B ∪ C) = (A ∪ B) ∪ C 
o A ∩ ∅ = ∅ 
A ∪ ∅ = A 
o A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) 
Ac 
x ∈ A ∩ B → x ∈ A ∧ x ∈ 
A ∩ B = B ∩ A 
B(por definición) → x ∈ B ∧ x ∈ A(conmutatividad del conectivo ∧) → x ∈ B ∩ 
A(por definición). 
A △ B = (A ∖ B) ∪ (B ∖ A) = (A ∪ B) ∖ (A ∩ B) 
{x ∈ A x⁄ ∉ B} 
• A △ B = B △ A 
• A △ (B △ C) = (A △ B) △ C 
A ∩ B ⊂ B ∩ A 
• A △ A = ∅ 
A △ ∅ = A 
• A ∩ (B △ C) = (A ∩ B) △ (A ∩ C) 
o A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) 
o A ∪ U = U 
A ∩ U = A 
o (A ∪ B)c = Ac ∩ Bc 
o (A ∩ B)c = Ac ∪ Bc 
A ∖ B = A ∩ Bc 
• A × (B ∪ C) = (A × B) ∪ (A × C) 
• A × (B ∩ C) = (A × B) ∩ (A × C) 
• (A × B) ∩ (C × D) = (A ∩ C) × (B ∩ D) 
• A × B = ∅,si y sólo si A = ∅ o B = ∅. 
A × B = {(a,b) a⁄ ∈ A y b ∈ B} 
A2 
R2 
A × A 
R 
R × R 
A × B ≠ B × A 
A × B ≠ B × A 
A ≠ B 
A ≠ B 
A ≠ B 
{b ∈ B ∕ ∃a ∈ 
A tal que (a,b) ∈ R} 
R ⊂ A × B 
(a,b) ∈ R 
a 
b 
b 
a 
a 
a 
{a ∃b ⁄ tal que (a,b) ∈ R} 
BR = {b ∃a ⁄ tal que (a,b) ∈ R} 
x ∈ 
A 
R 
A 
A 
(x,x) ∈ R 
(x,x) ∈ R 
(x,x) ∈ R 
Para todo x,y,z ∈ A,(x,y),(y,z) ∈ R ⇒ (x,z) ∈ R 
A = {1,2,3} 
x,y ∈ A 
R 
R 
A 
A 
R ⊂ A × A 
R ⊂ A × A 
(x,y) ∈ R ⇒ (y,x) ∈ R 
(x,y) ∈ R ⇒ (y,x) ∈ R 
12 = 24 
A1 ∪ A2 ∪ ⋯ = A y Ai ∩ Aj = ∅ para todo i ≠ j 
A1 = {1,3,5},A2 = {2,4,6},A3 = {7},A4 = 
{8,9} 
A = {1,2,3,4,5,6,7,8,9} 
A 
A 
aRb 
A = N = {0,1,2,3,...} 
a 
b 
b 
A = N 
aRb 
R 
a 
b 
(2,4) ∈ D pero (5,6) ∉ D 
A 
A 
A 
2 
2 
2 
6R8 
6R8 
6R8 
6R8 
3R7 
3R7 
3R7 
3R7 
3R7 
A 
A 
(2,6) ∈ D pero (6,2) ∉ D 
A 
R1 = 
{(1,a),(3,a),(1,c),(4,e)} 
R 
{b,c,d} 
{b,c,d} 
(1,a) 
(1,c) 
(1,c) 
(0,1) 
(0,−1) 
f 
f 
g = {x,y y⁄ = 2x;x,y ∈ R} 
g = {x,y y⁄ = 2x;x,y ∈ R} 
g = {x,y y⁄ = 2x;x,y ∈ R} 
A = R 
f = {x,y x2 ⁄ + y2 = 1;x,y ∈ R} 
f 
f 
A = {1,3,4,6} 
f 
A 
B = {a,b,c,d,e} 
A 
B 
B 
B 
R = {(1,b),(4,c),(3,d),(6,c)} 
R = {(1,b),(4,c),(3,d),(6,c)} 
(a,b) 
(a,c) 
{x ∈ B existe ⁄ x ∈ A tal que (a,b) ∈ f} 
f 
f 
b = c 
b = c 
b = c 
f(a) = b 
afb 
(a,b) ∈ f} 
(a,b) ∈ f} 
id 
f,g,h 
A 
(g ∘ f):A → C 
A = {a,b,c},C = {1,2,3},B = {x,y,z},f = {(a,1),(b,1),(c,2)},g = 
{(1,x),(2,y),(3,y)} 
f(x) = x2 
(g ∘ f)(x) = g(f(x)) = g(2x + 3) = 
(2x + 3)2 
(f ∘ g)(x) = f(g(x)) = 
f(x2) = 2x2 + 3 
(g ∘ f):R → R 
(g ∘ f):R → R 
(f ∘ g):R → R 
(f ∘ g):R → R 
(g ∘ f):A → B 
{(a,x),(b,x),(c,y)} 
{(a,x),(b,x),(c,y)} 
f:A → B y g:B → C 
f 
g 
g 
A = B = C = R 
h:A → C 
g 
B 
R 
f:R → R 
h(a) = g(f(a)) 
f(x) = 2x + 3 
f(x) = 2x + 3 
a ∈ A 
a ∈ A 
g:R → R 
g:R → R 
g:R → R 
A 
id(a) = a 
f:A → B 
f 
Im(F) 
a 
a 
f 
A 
A 
A 
A 
A 
idA 
idB(f(a)) = 
f(a) 
A 
idB 
idB 
B 
B 
B 
f:A → B 
f:A → B 
f:A → B 
f:A → B 
idB 
(idB ∘ f):A → B 
(idB ∘ f):A → B 
(idB ∘ f):A → B 
(idB ∘ f):A → B 
(idB ∘ f):A → B 
(f(a) = f(b)) ⇒ (a = b) 
2a + 
3 = 2b + 3 ⇒ 2a = 2b ⇒ a = b 
g(−1) = g(1) = 
1 
f(x) = 2x + 3 
g(x) = x2 
g(x) = x2 
f(a) = f(b) 
f(a) = f(b) 
h(a) = 1,h(b) = 2,h(c) = 3,h(c) = 
4 
h:C → D 
C = {a,b,c,d} 
C = {a,b,c,d} 
D = {1,2,3,4,5} 
D = {1,2,3,4,5} 
D = {1,2,3,4,5} 
k:C → E 
C 
C 
h 
h 
h 
E = {m,n} 
f(x) = 2x + 3 
k(a) = m,k(b) = n,k(c) = m,k(d) = n 
f:A → B 
k 
k 
g(x) = x2 
f 
f:A → B 
a,b ∈ A,a ≠ b entonces f(a) ≠ f(b) 
bR−1a 
f(a) = b 
b ∈ B 
(f−1 ∘ f)(a) = 
aRb 
f−1 
f 
f−1 
f−1 
f:A → B 
f:A → B 
f 
f 
(f−1 ∘ f):A → A 
(f−1 ∘ f):A → A 
(f−1 ∘ f):A → A 
(f−1)(f(a)) = f−1(b) = a 
(f ∘ f−1) = idB 
(f−1 ∘ f) = idA 
A 
In 
R 
f(x) = 2x + 3 
n 
n 
A 
B 
B 
R−1 
R−1 
R−1 
B 
B 
B 
B 
A 
A 
A 
A 
A 
f−1 
n 
f−1:B → A 
A 
In = {1,2,3,4,...,n} 
f−1(b) = a 
f−1(b) = a 
f 
{1,2,3,4,5} 
V = {a,e,i,o,u} 
n 
3 
In 
f 
f(n) = 3n 
Planteamiento de problemas 
• 
• 
• 
• 
Básica: