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-# COMP182 Homework 2
+# COMP182 Homework 2: Proofs, Sets, and Functions
 
 ## Student Responses
 
-<!-- Write your solutions below. Add sections for each problem as needed. -->
-<!-- Use proper mathematical notation where applicable. -->
+---
 
-[Your solutions here]
+## 1 Sets
 
+### Problem 1
+
+**(a)** Prove that (A △ B) △ B = A.
+
+**Proof.** We show both inclusions by element chasing.
+
+Recall that x ∈ X △ Y if and only if x is in exactly one of X and Y, i.e., (x ∈ X and x ∉ Y) or (x ∉ X and x ∈ Y).
+
+**(⊆)** Let x ∈ (A △ B) △ B. Then x is in exactly one of (A △ B) and B.
+
+- **Case 1:** x ∈ (A △ B) and x ∉ B. Since x ∈ A △ B, either x ∈ A \ B or x ∈ B \ A. Since x ∉ B, we must have x ∈ A \ B, so x ∈ A.
+- **Case 2:** x ∉ (A △ B) and x ∈ B. Since x ∉ A △ B, the element x is not in exactly one of A and B—it is either in both or in neither. Since x ∈ B, we must have x ∈ A ∩ B, so x ∈ A.
+
+In both cases, x ∈ A.
+
+**(⊇)** Let x ∈ A. We consider two sub-cases:
+
+- **Case 1:** x ∈ A and x ∉ B. Then x ∈ A \ B ⊆ A △ B. Since x ∉ B, we have x ∈ (A △ B) \ B ⊆ (A △ B) △ B.
+- **Case 2:** x ∈ A and x ∈ B. Then x ∈ A ∩ B, so x ∉ A △ B (since x is in both sets, not exactly one). Since x ∈ B and x ∉ A △ B, we have x ∈ B \ (A △ B) ⊆ (A △ B) △ B.
+
+In both cases, x ∈ (A △ B) △ B.
+
+Since both inclusions hold, (A △ B) △ B = A. ∎
+
+---
+
+**(b)** Is A △ (B △ C) = (A △ B) △ C? **Yes**, this is true.
+
+**Proof.** We show that x ∈ A △ (B △ C) if and only if x is in an odd number of the sets A, B, C, and likewise for (A △ B) △ C.
+
+Note that x ∈ X △ Y iff x is in exactly one of X, Y.
+
+For the left side, x ∈ A △ (B △ C) iff x is in exactly one of A and (B △ C). Enumerating all membership combinations for A, B, C:
+
+| x ∈ A? | x ∈ B? | x ∈ C? | x ∈ B △ C? | x ∈ A △ (B △ C)? |
+|---------|--------|--------|------------|-------------------|
+| Yes     | No     | No     | No         | Yes               |
+| Yes     | Yes    | No     | Yes        | No                |
+| Yes     | No     | Yes    | Yes        | No                |
+| Yes     | Yes    | Yes    | No         | Yes               |
+| No      | No     | No     | No         | No                |
+| No      | Yes    | No     | Yes        | Yes               |
+| No      | No     | Yes    | Yes        | Yes               |
+| No      | Yes    | Yes    | No         | No                |
+
+So x ∈ A △ (B △ C) iff x belongs to an odd number of the sets {A, B, C}.
+
+For the right side, x ∈ (A △ B) △ C iff x is in exactly one of (A △ B) and C:
+
+| x ∈ A? | x ∈ B? | x ∈ C? | x ∈ A △ B? | x ∈ (A △ B) △ C? |
+|---------|--------|--------|------------|-------------------|
+| Yes     | No     | No     | Yes        | Yes               |
+| Yes     | Yes    | No     | No         | No                |
+| Yes     | No     | Yes    | Yes        | No                |
+| Yes     | Yes    | Yes    | No         | Yes               |
+| No      | No     | No     | No         | No                |
+| No      | Yes    | No     | Yes        | Yes               |
+| No      | No     | Yes    | No         | Yes               |
+| No      | Yes    | Yes    | Yes        | No                |
+
+This is exactly the same set of rows marked "Yes"—precisely those where x belongs to an odd number of {A, B, C}.
+
+Since both sides consist of exactly those elements belonging to an odd number of A, B, C, we have A △ (B △ C) = (A △ B) △ C. ∎
+
+---
+
+### Problem 2
+
+**Prove that** |A ∩ B| ≤ |A ∪ B| for finite sets A, B.
+
+**Proof.** Since every element in A ∩ B is in both A and B, it is certainly in A ∪ B. Thus A ∩ B ⊆ A ∪ B. For finite sets, if S ⊆ T then |S| ≤ |T| (since an injective function from S to T is given by the inclusion map, and for finite sets an injection implies the domain is no larger than the codomain). Therefore |A ∩ B| ≤ |A ∪ B|. ∎
+
+**When is there equality?** We have |A ∩ B| = |A ∪ B| if and only if A = B.
+
+*Proof:* Since A ∩ B ⊆ A ∪ B, equality of cardinalities for finite sets implies A ∩ B = A ∪ B (a finite subset of a finite set with the same cardinality must equal the whole set). Now A ∩ B = A ∪ B implies:
+- A ⊆ A ∪ B = A ∩ B ⊆ A, so A = A ∩ B, meaning A ⊆ B.
+- Similarly B ⊆ A.
+- Therefore A = B.
+
+Conversely, if A = B, then A ∩ B = A = A ∪ B, so equality holds.
+
+---
+
+### Problem 3
+
+**Set-based formulation of ordered pairs.**
+
+We use the **Kuratowski definition**: define the ordered pair as
+
+> (a, b) := {{a}, {a, b}}.
+
+**Theorem.** (a, b) = (c, d) if and only if a = c and b = d.
+
+**Proof.**
+
+(⇐) If a = c and b = d, then {{a}, {a, b}} = {{c}, {c, d}} immediately.
+
+(⇒) Suppose {{a}, {a, b}} = {{c}, {c, d}}. We consider two cases.
+
+**Case 1: a = b.** Then (a, b) = {{a}, {a, a}} = {{a}, {a}} = {{a}}. So {{a}} = {{c}, {c, d}}. A set with one element equals a set with two listed elements only if those two elements are the same, so {c} = {c, d} = {a}. From {c} = {a} we get c = a. From {c, d} = {a} and c = a, we get d = a = b. Thus a = c and b = d.
+
+**Case 2: a ≠ b.** Then {a} ≠ {a, b} (since {a, b} has two distinct elements). The set {{a}, {a, b}} has exactly two elements. Since it equals {{c}, {c, d}}, which also has at most two elements, we consider which elements correspond:
+
+- **Sub-case 2a:** {a} = {c} and {a, b} = {c, d}. From {a} = {c}, we get a = c. Then {a, b} = {a, d}, which (since a ≠ b) gives b = d.
+
+- **Sub-case 2b:** {a} = {c, d} and {a, b} = {c}. From {a, b} = {c}, both a and b equal c, so a = b = c. But this contradicts a ≠ b.
+
+In all cases, a = c and b = d. ∎
+
+---
+
+## 2 Functions
+
+### Problem 1
+
+Let f(x) = x² + 1 and g(x) = x + 2.
+
+**f ∘ g:** (f ∘ g)(x) = f(g(x)) = f(x + 2) = (x + 2)² + 1 = x² + 4x + 4 + 1 = **x² + 4x + 5**.
+
+**g ∘ f:** (g ∘ f)(x) = g(f(x)) = g(x² + 1) = (x² + 1) + 2 = **x² + 3**.
+
+---
+
+### Problem 2
+
+**Claim:** If f and f ∘ g are one-to-one, then g is one-to-one.
+
+**Proof.** Suppose g(x₁) = g(x₂) for some x₁, x₂ in the domain of g. Then:
+
+f(g(x₁)) = f(g(x₂))
+
+which means (f ∘ g)(x₁) = (f ∘ g)(x₂). Since f ∘ g is one-to-one, this implies x₁ = x₂.
+
+Therefore g is one-to-one. ∎
+
+---
+
+### Problem 3
+
+Let f : A → B and S, T ⊆ A.
+
+**(a)** Prove that f(S ∪ T) = f(S) ∪ f(T).
+
+**Proof.**
+
+(⊆) Let y ∈ f(S ∪ T). Then y = f(x) for some x ∈ S ∪ T. Either x ∈ S or x ∈ T (or both).
+- If x ∈ S, then y = f(x) ∈ f(S).
+- If x ∈ T, then y = f(x) ∈ f(T).
+
+In either case, y ∈ f(S) ∪ f(T).
+
+(⊇) Let y ∈ f(S) ∪ f(T). Then y ∈ f(S) or y ∈ f(T).
+- If y ∈ f(S), then y = f(x) for some x ∈ S. Since S ⊆ S ∪ T, we have x ∈ S ∪ T, so y ∈ f(S ∪ T).
+- If y ∈ f(T), then y = f(x) for some x ∈ T ⊆ S ∪ T, so y ∈ f(S ∪ T).
+
+Since both inclusions hold, f(S ∪ T) = f(S) ∪ f(T). ∎
+
+---
+
+**(b)** Prove that f(S ∩ T) ⊆ f(S) ∩ f(T).
+
+**Proof.** Let y ∈ f(S ∩ T). Then y = f(x) for some x ∈ S ∩ T. Since x ∈ S ∩ T, we have x ∈ S and x ∈ T. Therefore:
+- y = f(x) ∈ f(S) (since x ∈ S), and
+- y = f(x) ∈ f(T) (since x ∈ T).
+
+So y ∈ f(S) ∩ f(T). This proves f(S ∩ T) ⊆ f(S) ∩ f(T). ∎
+
+*Note:* Equality does not hold in general. For a counterexample, let A = {1, 2}, B = {3}, f(1) = f(2) = 3, S = {1}, T = {2}. Then S ∩ T = ∅, so f(S ∩ T) = ∅, but f(S) ∩ f(T) = {3} ∩ {3} = {3} ≠ ∅.
+
+---
+
+### Problem 4
+
+**Claim:** There exists a bijection h : B → C.
+
+**Proof.** Since f : A → B is a bijection, it has an inverse f⁻¹ : B → A, which is also a bijection. Define
+
+> h = g ∘ f⁻¹ : B → C.
+
+We show h is a bijection by verifying it is both injective and surjective.
+
+**Injective:** Suppose h(b₁) = h(b₂) for b₁, b₂ ∈ B. Then g(f⁻¹(b₁)) = g(f⁻¹(b₂)). Since g is injective (as it is a bijection), f⁻¹(b₁) = f⁻¹(b₂). Since f⁻¹ is injective (as it is a bijection), b₁ = b₂. So h is injective.
+
+**Surjective:** Let c ∈ C. Since g is surjective, there exists a ∈ A with g(a) = c. Let b = f(a) ∈ B. Then f⁻¹(b) = a, so h(b) = g(f⁻¹(b)) = g(a) = c. So h is surjective.
+
+Since h is both injective and surjective, h is a bijection from B to C. ∎
+
+---
+
+### Problem 5
+
+**Claim:** If A and B are finite sets with |A| = |B| and f : A → B is injective, then f is surjective.
+
+**Proof.** Since f is injective, distinct elements of A map to distinct elements of B. Therefore the image f(A) = {f(a) : a ∈ A} has exactly |A| distinct elements, i.e., |f(A)| = |A|.
+
+Since f(A) ⊆ B and both are finite with |f(A)| = |A| = |B|, the subset f(A) must equal all of B (a finite set cannot have a proper subset of the same cardinality). Therefore f(A) = B, which means f is surjective. ∎
+
+---
+
+### Problem 6
+
+**Claim:** No, an injective function f : A → B between infinite sets with a bijection between them is **not** necessarily surjective.
+
+**Counterexample.** Let A = B = ℕ = {0, 1, 2, 3, ...} (the natural numbers). The identity function id : ℕ → ℕ is a bijection, so a bijection between A and B exists.
+
+Define f : ℕ → ℕ by f(n) = 2n.
+
+- **f is injective:** If f(n₁) = f(n₂), then 2n₁ = 2n₂, so n₁ = n₂.
+- **f is not surjective:** The element 1 ∈ ℕ is not in the image of f, since 2n = 1 has no solution in ℕ.
+
+Therefore, the existence of a bijection between two infinite sets does not guarantee that every injection between them is a surjection. ∎
+
+---
+
+## 3 Relations
+
+### Problem 1
+
+**Claim:** R₁ defined on ordered pairs of positive integers by (a, b) R₁ (c, d) iff ad = bc is an equivalence relation.
+
+**Proof.**
+
+**Reflexive:** For any (a, b), we need (a, b) R₁ (a, b), i.e., ab = ba. This holds since multiplication of integers is commutative.
+
+**Symmetric:** Suppose (a, b) R₁ (c, d), so ad = bc. Then cb = da (the same equation rearranged), which means (c, d) R₁ (a, b).
+
+**Transitive:** Suppose (a, b) R₁ (c, d) and (c, d) R₁ (e, f). Then:
+- ad = bc ... (1)
+- cf = de ... (2)
+
+We need to show af = be.
+
+Multiply (1) by f: adf = bcf.
+Multiply (2) by b: bcf = bde.
+
+Therefore adf = bde. Since d is a positive integer (d > 0), we can divide both sides by d:
+
+af = be.
+
+This gives (a, b) R₁ (e, f). Therefore R₁ is transitive.
+
+Since R₁ is reflexive, symmetric, and transitive, it is an equivalence relation. ∎
+
+*Remark:* The equivalence class of (a, b) corresponds to the rational number a/b, and the relation captures exactly when two fractions are equal: a/b = c/d iff ad = bc.
+
+---
+
+### Problem 2
+
+**Claim:** If R₁ and R₂ are equivalence relations on a set A, then R₁ ∩ R₂ is an equivalence relation on A.
+
+**Proof.**
+
+**Reflexive:** For all a ∈ A, since R₁ is reflexive, (a, a) ∈ R₁, and since R₂ is reflexive, (a, a) ∈ R₂. Therefore (a, a) ∈ R₁ ∩ R₂.
+
+**Symmetric:** Suppose (a, b) ∈ R₁ ∩ R₂. Then (a, b) ∈ R₁ and (a, b) ∈ R₂. By symmetry of R₁, (b, a) ∈ R₁. By symmetry of R₂, (b, a) ∈ R₂. Therefore (b, a) ∈ R₁ ∩ R₂.
+
+**Transitive:** Suppose (a, b) ∈ R₁ ∩ R₂ and (b, c) ∈ R₁ ∩ R₂. Then:
+- (a, b) ∈ R₁ and (b, c) ∈ R₁, so by transitivity of R₁, (a, c) ∈ R₁.
+- (a, b) ∈ R₂ and (b, c) ∈ R₂, so by transitivity of R₂, (a, c) ∈ R₂.
+
+Therefore (a, c) ∈ R₁ ∩ R₂.
+
+Since R₁ ∩ R₂ is reflexive, symmetric, and transitive, it is an equivalence relation. ∎
+
+---
+
+### Problem 3
+
+**Definition.** A relation R on A is *circular* if (a, b) ∈ R and (b, c) ∈ R implies (c, a) ∈ R.
+
+**Claim:** A relation R on A is an equivalence relation if and only if it is reflexive and circular.
+
+**Proof.**
+
+**(⇒)** Suppose R is an equivalence relation (reflexive, symmetric, transitive). Clearly R is reflexive. To show R is circular: suppose (a, b) ∈ R and (b, c) ∈ R. By transitivity, (a, c) ∈ R. By symmetry, (c, a) ∈ R. So R is circular.
+
+**(⇐)** Suppose R is reflexive and circular. We show R is symmetric and transitive.
+
+**Symmetric:** Suppose (a, b) ∈ R. Since R is reflexive, (b, b) ∈ R. Applying the circular property to (a, b) ∈ R and (b, b) ∈ R, we get (b, a) ∈ R. So R is symmetric.
+
+**Transitive:** Suppose (a, b) ∈ R and (b, c) ∈ R. By the circular property, (c, a) ∈ R. By symmetry (proved above), (a, c) ∈ R. So R is transitive.
+
+Since R is reflexive, symmetric, and transitive, R is an equivalence relation. ∎
+
+---
+
+### Problem 4
+
+**(a)** A poset with a minimal element but no maximal element.
+
+**Example:** (ℕ, ≤) where ℕ = {0, 1, 2, 3, ...} with the usual ordering.
+
+- **Minimal element:** 0 is minimal because there is no n ∈ ℕ with n < 0.
+- **No maximal element:** For any n ∈ ℕ, we have n < n + 1, so n is not maximal.
+
+---
+
+**(b)** A poset with neither a maximal nor a minimal element.
+
+**Example:** (ℤ, ≤) where ℤ = {..., -2, -1, 0, 1, 2, ...} with the usual ordering.
+
+- **No minimal element:** For any n ∈ ℤ, we have n - 1 < n, so n is not minimal.
+- **No maximal element:** For any n ∈ ℤ, we have n < n + 1, so n is not maximal.
+
+---
+
+### Problem 5
+
+**Can a non-empty finite partially ordered set have no minimal element?**
+
+**No.**
+
+**Proof.** Let (A, ≤) be a non-empty finite partially ordered set. Suppose for contradiction that A has no minimal element. Pick any element a₁ ∈ A. Since a₁ is not minimal, there exists a₂ ∈ A with a₂ < a₁ (i.e., a₂ ≤ a₁ and a₂ ≠ a₁). Since a₂ is not minimal, there exists a₃ ∈ A with a₃ < a₂. Continuing inductively, we obtain an infinite sequence
+
+a₁ > a₂ > a₃ > a₄ > ⋯
+
+All elements in this sequence are distinct: if i < j, then by transitivity of ≤ we have aⱼ ≤ aⱼ₋₁ ≤ ⋯ ≤ aᵢ₊₁ ≤ aᵢ, so aⱼ ≤ aᵢ. Also aⱼ ≠ aᵢ (since each step is strict: aₖ₊₁ < aₖ, and if aⱼ = aᵢ, then aᵢ ≤ aⱼ = aᵢ₊₁ ≤ ... ≤ aᵢ₊₁, which combined with aᵢ₊₁ < aᵢ gives aᵢ ≤ aᵢ₊₁ and aᵢ₊₁ < aᵢ, a contradiction).
+
+So the sequence a₁, a₂, a₃, ... consists of infinitely many distinct elements of A. But A is finite, so it cannot contain infinitely many distinct elements. This is a contradiction.
+
+Therefore, every non-empty finite partially ordered set must have a minimal element, and A cannot be finite. ∎