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Lista de Exercícios Análise Matemática, Exercícios de Análise Matemática

Referente ao capítulo 1 do livro Análise de Lages

Tipologia: Exercícios

2023

À venda por 12/01/2023

Carolina.Prado
Carolina.Prado 🇧🇷

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bg1
Mathematics - 2023
Insper
PhD Program in Business Economics
Problem Set 1
1. Show that for all statements p, q, q1eq2:
(a) p ¬ (¬p).
(b) ¬(pq)(¬p)(¬q).
(c) ¬(pq)(¬p)(¬q).
(d) p(q1q2)(pq1)(pq2).
(e) p(q1q2)(pq1)(pq2).
(f) p(q1q2)(pq1)q2.1
2. Given two statements pand q, is it always true that2
q(pq)?
3. Show that the compound propositions P in the following items are tau-
tologies:
(a) P:p(pq)
(b) P: (pq)p
(c) P: [(pq)(qq1)] [pq1]. 3
(d) P: (pq)(pq)
1We note that this property implies that given A,B, C 2X, where X=, it follows that:
(AB) C=A (BC).
2A statement that is always true is called a tautology. For instance, given p, the statement
[p ¬p] is a tautology (note that we cannot combine pwith ¬pby taking T(F) and T(F) at
the same time):
p¬p[p ¬p]
T F T
F T T
In many cases, a tautology is an expression or phrase that says the same thing twice, just
in a different way. In special, the equivalences found in the items (a) to (f) in the previous
exercise are tautologies.
On the other hand, a statement is called a contradiction if it is always false. For instance,
given p, the statement [p ¬p] is a contradiction:
p¬p[p ¬p]
T F F
F T F
A statement that assumes values T and F is called a contingency. For instance, given pand
q, by what we already see, [pq],[pq],[pq], and [pq] are examples of contingencies.
3This is called the law of syllogism or reasoning by transitivity
1
pf3
pf4

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Mathematics - 2023 Insper PhD Program in Business Economics Problem Set 1

  1. Show that for all statements p, q, q 1 e q 2 : (a) p ⇔ ¬ (¬p). (b) ¬ (p ∧ q) ⇔ (¬p) ∨ (¬q). (c) ¬ (p ∨ q) ⇔ (¬p) ∧ (¬q). (d) p ∨ (q 1 ∧ q 2 ) ⇔ (p ∨ q 1 ) ∧ (p ∨ q 2 ). (e) p ∧ (q 1 ∨ q 2 ) ⇔ (p ∧ q 1 ) ∨ (p ∧ q 2 ). (f) p ⊻ (q 1 ⊻ q 2 ) ⇔ (p ⊻ q 1 ) ⊻ q 2. 1
  2. Given two statements p and q, is it always true that^2 q ⇒ (p ⇒ q)?
  3. Show that the compound propositions P in the following items are tau- tologies: (a) P : p ⇒ (p ∨ q) (b) P : (p ∧ q) ⇒ p (c) P : [(p ⇒ q) ∧ (q ⇒ q 1 )] ⇒ [p ⇒ q 1 ]. 3 (d) P : (p ∧ q) ⇒ (p ∨ q) (^1) We note that this property implies that given A, B, C ∈ 2 X (^) , where X ̸= ∅, it follows that: (A∆B) ∆C = A∆ (B∆C). (^2) A statement that is always true is called a tautology. For instance, given p, the statement [p ∨ ¬p] is a tautology (note that we cannot combine p with ¬p by taking T (F ) and T (F ) at the same time): p ¬p [p ∨ ¬p] T F T F T T In many cases, a tautology is an expression or phrase that says the same thing twice, just in a different way. In special, the equivalences found in the items (a) to (f) in the previous exercise are tautologies. On the other hand, a statement is called a contradiction if it is always false. For instance, given p, the statement [p ∧ ¬p] is a contradiction: p ¬p [p ∧ ¬p] T F F F T F A statement that assumes values T and F is called a contingency. For instance, given p and q, by what we already see, [p ∧ q], [p ∨ q], [p ⇒ q], and [p ⇔ q] are examples of contingencies. (^3) This is called the law of syllogism or reasoning by transitivity

(e) P : (p ⇔ q) ⇔ [(p ⇒ q) ∧ (q ⇒ p)] (f) P : (p ⇔ q) ⇔ (¬p ⇔ ¬q) (g) P : (p ⇒ q) ⇔ (¬p ∨ q) (h) P : ¬ (p ⇒ q) ⇔ (p ∧ ¬q) (i) P : [p ∧ (p ⇒ q)] ⇒ q

  1. Consider a nonempty set X. Take A, B ∈ 2 X^ and let C ∈ 2 X^ that satisfies the following properties: (i) A ⊆ C and B ⊆ C, (ii) For all D ∈ 2 X^ , A ⊆ D ∧ B ⊆ D ⇒ C ⊆ D. Show that C = A ∪ B. (Thus, we can conclude that A ∪ B is the smallest set containing both A and B).
  2. State and prove the analogous result characterizing A ∩ B.
  3. Show that for all A, A′, B, B′^ ∈ 2 X^ such that A ⊆ B and A′^ ⊆ B′^ :

(i) A ∪ A′^ ⊆ B ∪ B′; (ii) A ∩ A′^ ⊆ B ∩ B′; (iii) A × A′^ ⊆ B × B′

  1. Given A, B ∈ 2 X^ , where X is a nonempty set, show that:

(a) A ∩ B = ∅ ⇔ A ⊆ Bc; (b) A ∪ B = X ⇔ Ac^ ⊆ B; (c) A ⊆ B ⇔ A ∩ Bc^ = ∅; (d) (A ∩ B = ∅ ∧ A ∪ B = X) ⇔ B = Ac; (e) B ⊆ A ⇒ ∀C ∈ 2 X^ , A ∩ (B ∪ C) = (A ∩ C) ∪ B; (f) ∃C ∈ 2 X^ s.t. A ∩ (B ∪ C) = (A ∩ C) ∪ B ⇒ B ⊆ A; (g) A = B ⇔ (A ∩ Bc) ∪ (Ac^ ∩ B) = ∅; Now, consider also C ∈ 2 Y^ , where Y is a nonempty set, and show the following properties about subsets of X×Y = {(x, y) : x ∈ X and y ∈ Y } : (h) (A ∪ B) × C = (A × C) ∪ (B × C) ; (i) (A ∩ B) × C = (A × C) ∩ (B × C) ; (j) (A\B) × C = (A × C) \ (B × C) ;

  1. Find an example of sets A, B, and C such that

(A ∪ B) ∩ C ̸= A ∪ (B ∩ C).

(b) We say that g : Y → X is a right inverse of f when f og = IdY : Y → Y, i.e., f (g (y)) = y for all y ∈ Y. Show that there exists a right inverse for f if and only if f is onto. (c) Compare the results from (a) and (b) in this question with the def- inition the inverse function f −^1 for a bijective function f : X → Y. Also, show that if f : X → Y and g : Y → Z are bijections then gof has a inverse (gof )−^1 : Z → X with (gof )−^1 = f −^1 og−^1.

  1. Let f : X → Y and g : Y → Z two functions. Show that: (a) If g ◦ f is one-to-one then f is one-to-one; (b) If g ◦ f is onto then g is onto; (c) If g ◦ f is one-to-one and f is onto then g is one-to-one; (d) If g ◦ f is onto and g is one-to-one then f is onto;
  2. Let f : X → X be a function such that f ◦ f ◦ f = f. Show that f is one-to-one if and only if f is onto.
  3. Let X a nonempty set. Show that for all f : X → 2 X^ , the function f cannot be onto.
  4. Given a nonempty set X, let R be a relation on X. For each x ∈ X, denote its equivalence class by Ex := {y ∈ X : yRx}. Show that R is an equivalence relation on X if and only if the family of sets Λ := {Ex : x ∈ X} is a partition of X.^5
  5. Given a function f : A → B, consider the binary relation defined by xRy ⇐⇒ f (x) = f (y). Show that R ⊆ A^2 is an equivalence relation and that for each x ∈ A, the corresponding equivalence class Ex satisfies Ex = f −^1 ({f (x)}). (^5) A partition of X is any collection {Aλ}λ∈Λ (Λ is a set of index) of nonempty subsets of X such that (^) [

λ∈Λ

Aλ = X,

and for all λ 1 ̸= λ 2 , Aλ 1 ∩ Aλ 2 = ∅.