Preserves inclusions, unions intersections

[tomboi03]

Let f: A-> B and let A_i$$\subset$$ A and B_i$$\subset$$B for i=0 and i=1. Shwo that f^-1 preserves inclusions, unions, intersections, and differences of sets:
a. B₀$$\subset$$B₁ = f^-1(B₀)$$\subset$$f^-1(B₁)
b. f^-1(B₀$$\cup$$B₁) = f^-1(B₀)$$\cup$$f^-1(B₁)
c. f^-1(B₀$$\cap$$B₁) = f^-1(B₀)$$\cap$$f^-1(B₁)
d. f^-1(B₀-B₁) = f^-1(B₀)-f^-1(B₁)

Show that f preserves inclusions and unions only:
e. A₀$$\subset$$A1₁ => f(A₀)$$\subset$$f(A₁)
f. f(A₀$$\cup$$A₁)=f(A₀)$$\cup$$f(A₁)
g.f(A₀$$\cap$$A₁)=f(A₀)$$\cap$$f(A₁); show that equality holds if f is injective
h.f(A₀-A₁)=f(A₀)-f(A₁); show that equality holds if is injective

Thanks

 

[jvicens]

for your post! I am happy to help clarify and provide proof for the statements made in the forum post.

First, let's define some terms for clarity:
- f-1: the inverse of function f, mapping from B to A
- A0, A1: subsets of A
- B0, B1: subsets of B

Now, let's prove each statement one by one:

a. B0\subsetB1 = f-1(B0)\subsetf-1(B1)
Proof: Let x∈f-1(B0). This means that f(x)∈B0. Since B0\subsetB1, this also means that f(x)∈B1. Therefore, x∈f-1(B1). This shows that f-1(B0)\subsetf-1(B1).

b. f-1(B0\cupB1) = f-1(B0)\cupf-1(B1)
Proof: Let x∈f-1(B0\cupB1). This means that f(x)∈B0\cupB1. By definition of union, this means that f(x)∈B0 or f(x)∈B1. Therefore, x∈f-1(B0) or x∈f-1(B1). This shows that x∈f-1(B0)\cupf-1(B1). This also holds in the other direction, showing that f-1(B0\cupB1) = f-1(B0)\cupf-1(B1).

c. f-1(B0\capB1) = f-1(B0)\capf-1(B1)
Proof: Let x∈f-1(B0\capB1). This means that f(x)∈B0\capB1. By definition of intersection, this means that f(x)∈B0 and f(x)∈B1. Therefore, x∈f-1(B0) and x∈f-1(B1). This shows that x∈f-1(B0)\capf-1(B1). This also holds in the other direction, showing that f-1(B0\capB1) = f-1(B0)\capf-1(B1).

d. f-1(B0-B1) = f