Question 1189248: Let X and Y be a non-empty sets and f: X →Y
(i.) If {E𝛼}𝛼 ∈ 𝐴 is collection of subset of X, then
f(∪E𝛼) = ∪ 𝑓(E𝛼) and f(∩E𝛼) = ∩ 𝑓(E𝛼)
(ii.) If B and C are subset of X, then
f(C \ B) ⊇ f(C) \ f(B)
(iii.) If {E𝛼}𝛼 ∈ 𝐴 is collection of subset of X, then
f^-1(∪E𝛼) = ∪ f^-1(E𝛼) and f^-1(∩E𝛼) = ∩ f^-1(E𝛼)
(iv.) If B and C are subset of X, then
f^-1(C \ B) = f^-1(C) \ f^-1(B)
v. If E ⊆ f(X) and f(f^-1(E))=E, but if E⊂ X, then f^-1(f(E)) ⊇ 𝐸
Show The proof
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Let's prove the properties of functions and inverse functions you've listed.
**(i.) f(∪E𝛼) = ∪ f(E𝛼) and f(∩E𝛼) ⊆ ∩ f(E𝛼)**
* **f(∪E𝛼) = ∪ f(E𝛼):**
* (⊆) Let y ∈ f(∪E𝛼). This means there exists an x ∈ ∪E𝛼 such that f(x) = y. Since x ∈ ∪E𝛼, x ∈ E𝛼 for some α. Therefore, y = f(x) ∈ f(E𝛼) for some α. This implies y ∈ ∪ f(E𝛼).
* (⊇) Let y ∈ ∪ f(E𝛼). This means y ∈ f(E𝛼) for some α. Thus, there exists an x ∈ E𝛼 such that f(x) = y. Since x ∈ E𝛼, x ∈ ∪E𝛼. Therefore, y = f(x) ∈ f(∪E𝛼).
* **f(∩E𝛼) ⊆ ∩ f(E𝛼):**
Let y ∈ f(∩E𝛼). Then there exists an x ∈ ∩E𝛼 such that f(x) = y. Since x ∈ ∩E𝛼, x ∈ E𝛼 for all α. Therefore, y = f(x) ∈ f(E𝛼) for all α. This means y ∈ ∩ f(E𝛼).
*It's important to note that f(∩E𝛼) = ∩ f(E𝛼) does not necessarily hold. The inclusion can be proper.* Consider a function f(x) = x² and sets E₁ = {-1} and E₂ = {1}. Then E₁∩E₂ = ∅, so f(E₁∩E₂) = ∅. However, f(E₁) = {1} and f(E₂) = {1}, so f(E₁)∩f(E₂) = {1}. Thus, f(E₁∩E₂) ≠ f(E₁)∩f(E₂).
**(iii.) f⁻¹(∪E𝛼) = ∪ f⁻¹(E𝛼) and f⁻¹(∩E𝛼) = ∩ f⁻¹(E𝛼)**
These properties hold for inverse images, regardless of whether the function is invertible.
* **f⁻¹(∪E𝛼) = ∪ f⁻¹(E𝛼):**
* (⊆) Let x ∈ f⁻¹(∪E𝛼). This means f(x) ∈ ∪E𝛼. So, f(x) ∈ E𝛼 for some α. This implies x ∈ f⁻¹(E𝛼) for some α, which means x ∈ ∪ f⁻¹(E𝛼).
* (⊇) Let x ∈ ∪ f⁻¹(E𝛼). This means x ∈ f⁻¹(E𝛼) for some α. So, f(x) ∈ E𝛼 for some α. This implies f(x) ∈ ∪E𝛼, which means x ∈ f⁻¹(∪E𝛼).
* **f⁻¹(∩E𝛼) = ∩ f⁻¹(E𝛼):**
* (⊆) Let x ∈ f⁻¹(∩E𝛼). This means f(x) ∈ ∩E𝛼. So, f(x) ∈ E𝛼 for all α. This implies x ∈ f⁻¹(E𝛼) for all α, which means x ∈ ∩ f⁻¹(E𝛼).
* (⊇) Let x ∈ ∩ f⁻¹(E𝛼). This means x ∈ f⁻¹(E𝛼) for all α. So, f(x) ∈ E𝛼 for all α. This implies f(x) ∈ ∩E𝛼, which means x ∈ f⁻¹(∩E𝛼).
**(ii.) f(C \ B) ⊇ f(C) \ f(B)**
Let y ∈ f(C) \ f(B). This means y ∈ f(C) and y ∉ f(B). Since y ∈ f(C), there exists an x ∈ C such that f(x) = y. Since y ∉ f(B), there is no x' ∈ B such that f(x') = y. Since f(x) = y and y ∉ f(B), x cannot be in B. Therefore, x ∈ C \ B, which means y = f(x) ∈ f(C \ B).
*The inclusion can be proper.* Consider a function f(x) = c (a constant). Let C = {1, 2} and B = {1}. Then C \ B = {2}, so f(C \ B) = {c}. However, f(C) = {c} and f(B) = {c}, so f(C) \ f(B) = ∅. Thus, f(C \ B) ⊃ f(C) \ f(B).
**(iv.) f⁻¹(C \ B) = f⁻¹(C) \ f⁻¹(B)**
* (⊆) Let x ∈ f⁻¹(C \ B). This means f(x) ∈ C \ B. So, f(x) ∈ C and f(x) ∉ B. This implies x ∈ f⁻¹(C) and x ∉ f⁻¹(B). Therefore, x ∈ f⁻¹(C) \ f⁻¹(B).
* (⊇) Let x ∈ f⁻¹(C) \ f⁻¹(B). This means x ∈ f⁻¹(C) and x ∉ f⁻¹(B). So, f(x) ∈ C and f(x) ∉ B. This implies f(x) ∈ C \ B, which means x ∈ f⁻¹(C \ B).
**(v.) If E ⊆ f(X) then f(f⁻¹(E)) = E, but if E ⊂ X, then f⁻¹(f(E)) ⊇ E**
* If E ⊆ f(X), then f(f⁻¹(E)) = E:
Let y ∈ E. Since E ⊆ f(X), there exists some x ∈ X such that f(x) = y. This means x ∈ f⁻¹(E). Therefore, f(x) = y ∈ f(f⁻¹(E)). So E ⊆ f(f⁻¹(E)).
Conversely, let y ∈ f(f⁻¹(E)). Then there exists x ∈ f⁻¹(E) such that f(x) = y. Since x ∈ f⁻¹(E), f(x) = y ∈ E. Therefore f(f⁻¹(E)) ⊆ E.
Hence f(f⁻¹(E)) = E
* If E ⊂ X, then f⁻¹(f(E)) ⊇ E:
Let x ∈ E. Then f(x) ∈ f(E), which means x ∈ f⁻¹(f(E)). Therefore E ⊆ f⁻¹(f(E)).
However, it's important to note that f⁻¹(f(E)) = E does not necessarily hold when E ⊂ X. The inclusion can be proper.
Consider a function f: {1, 2} → {a} defined by f(1) = a and f(2) = a. Let E = {1}. Then f(E) = {a}, and f⁻¹(f(E)) = f⁻¹({a}) = {1, 2}. Thus, E ⊂ f⁻¹(f(E)).
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