Question

Given a function I and a subset A of its domain, let I(A) represent the range of lover the set A; that is, I(A) = {I(x) : x E A}. 12 Chapter 1. The Real Numbers (a) Let f(x) = x2. If A = [0,2] (the closed interval {x E R : ° ~ x ~ 2}) and B = [1,4]' find f(A) and f(B). Does f(A n B) = f(A) n f(B) in this case? Does f(A U B) = f(A) U f(B)? (b) Find two sets A and B for which f(A n B) =1= f(A) n f(B). (c) Show that, for an arbitrary function g : R ---+ R, it is always true that g(A n B) ~ g(A) n g(B) for all sets A, B ~ R. (d) Form and prove a conjecture about the relationship between g(A U B) and g(A) U g(B) for an arbitrary function g.

Answer 1

Step-by-step explanation:

(a)

Using the definition given from the problem

$f(A) = \{x^2 \, : \, x \in [0,2]\} = [0,4]\\f(B) = \{x^2 \, : \, x \in [1,4]\} = [1,16]\\f(A) \cap f(B) = [1,4] = f(A \cap B)$

Therefore it is true for intersection. Now for union, we have that

$A \cup B = [0,4]\\f(A\cup B ) = [0,16]\\f(A) = [0,4]\\f(B)= [1,16]\\f(A) \cup f(B) = [0,16]$

Therefore, for this case, it would be true that $f(A\cup B) = f(A)\cup f(B)$.

(b)

1 is not a set.

(c)

To begin with  

$A\cap B \subset A,B$

Therefore

$g(A\cap B) \subset g(A) \cap g(B)$

Now, given an element of $g(A) \cap g(B)$ it will belong to both sets, therefore it also belongs to $g(A\cap B)$, and you would have that

$g(A)\cap g(B) \subset g(A \cap B)$, therefore  $g(A)\cap g(B) = g(A \cap B)$.

(d)

To begin with $A,B \subset A \cup B$, therefore

$g(A) \cup g(b) \subset g(A\cup B)$