Let A and B be non-empty sets. Let S be a subset of A and T be a subset of B; an

Question

Let A and B be non-empty sets. Let S be a subset of A and T  be a subset of B; and let f : A --> B be a function.

Recall:

The image of S = f(S) = {f(x) : x is an element of S}:

The preimage of T = f^(-1)(T) = {x is an element of A: f(x) is an element of T}.

a) Let Y,Z be subsets of A. Prove f(Y union Z) = f(Y) union f(Z).

b) Let Y,Z be subsets of A. Find a counterexample that shows f(Y intersect Z) is not equal to f(Y) intersect f(Z).

c) Let P,Q be subsets of B. Determine if f^(-1)(P intersect Q) = f^(-1)(P) intersect  f^(-1)(Q) and prove your result.

Explanation / Answer

(a)

let x belongs to Y union Z, then f(x) belongs to f(Y union Z),

x belongs to Y union Z => x belongs to Y or Z

if x belongs to Y , then f(x) belongs to f(Y),

if x belongs to Z, then f(x)belongs to f(Z)

=> f(x) belongs to f(Y) or f(Z) => f(x) belongs to f(Y) union f(Z).

=>

f(Y union Z) is a subcset of f(Y) union f(Z)

now let f(x) belongs to f(Y) union f(Z)

=>f(x) belongs to f(Y) or f(Z)

=>

if f(x) belongs to f(Y), then x belongs to Y,

if f(x)belongs to f(Z), then x belongs to Z

=>

x belongs to Y or Z => x belongs to Y union Z => f(x) belongs to f(Y union Z)

=> f(Y) union f(Z) is a subset of f(Y union Z)

=>f(Y union Z) = f(Y) union f(Z)

thus proved

(b)

let A = {1,2,3}, B = {1,2,3}

let f(1) = 1. f(2) = 2, f(3) =1

let Y = {1,2}, Z = {2,3}

Y intersection Z = {2}

f(Y intersection Z) ={f(2)} = {2}

f(Y) = {1,2}

f(Z) = {1,2}

=>

f(Y) intersection f(Z) = {1,2}

thus f(Y intersection Z) is not equal to f(Y) intersection f(Z)

(c)

f^(-1)(P intersect Q) = f^(-1)(P) intersect f^(-1)(Q)

<=>

(P intersect Q) = f(f^(-1)(P) intersect f^(-1)(Q))

but from our above counter example f(Y intersection Z) is not equal to f(Y) intersection f(Z) =>

RHS may not be equal to f(f^(-1)(P) intersect f^(-1)(Q)) = P intersect Q

=> (P intersect Q) is not equal to f(f^(-1)(P) intersect f^(-1)(Q))

thus proved

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