Example 3.4.1

1. Let X be a set of the first-year students and let f : S → N be function which to every student assign the number of exercise-group to which he belongs. Assume, that there are 20 groups numbered from 101 to 120. The image of S under the function f is a 20-elements set {101, 102, ..., 120}.
2. Let g: R → R, g(x) = sin(x) and A = [-pi, - pi/2], B = [pi/2, pi]. Then  g(A) = [-1,0] i g(B) = [0, 1].
3. Let h : R^{+ →} R, h(x) = lg x (R⁺ denotes the set of all positive real numbers x, x>0) and let A = [0, 4], B= [2, 8]. Then h(A) = (- ∝, 2] i h(B) = [1, 3], see Fig. 3.4.1(b).

Fig. 3.4.1 The image of a set under a function.

Proof.

Ad 1. First of the equalities is rather obvious:

y ∈ f(A ∪ B) iff y = f(x) for an x ∈ A ∪ B iff x∈ A and y = f(x) or x ∈ B and y =f(x) iff y ∈ f(A) or y ∈ f(B) iff y ∈ f(A) ∪ f(B).

Ad 2. Proof of the second part is analogous. If y ∈ f(A ∩ B), then y = f(x) for an x ∈ A ∩ B. Hence,  x ∈ A and x ∈ B and also y = f(x), so y ∈ f(A) and y ∈ f(B) what is equivalent to y ∈ f(A) ∩ f(B).

Let's wonder why in the case of the second formula there is no equality. For that purpose we should analyse the examples once again. For the function g of the example  3.4.1(2), g(A ∩ B) is an empty set, and g(A) ∩ g(B) = {0}. However for the function h of the example  (3) we have: A ∩ B = [2, 4], h(A ∩ B) = [1, 2] = (- ∝ , 2] ∩ [1, 3] = [1, 2], compare Fig. 3.4.1(b).

Question 3.4.1 What condition has to be fulfilled by the function f so that  f(A ∩ B) = f(A) ∩ f(B) for any sets A and B?

Fig. 3.4.2 Inverse image of set under a function.

The figure 3.4.2 shows the inverse image of the set [3, 5] under function |x|. Inverse images under functions from the example 3.4.1 are noted below:

1. f ^-1({101,102}) = all student of the 1-st year, who belong to the exercise groups 101 and 102.
   
2. g(x) = sin x, g ^-1({0}) = {k*pi : k ∈ Z}, g^-1 ([-1,1]) = R .
3. h(x) = lg x, h ^-1([2,5]) = {x ∈ R : 4 ≤ x ≤ 32}.

Proof is very simmilar in both cases. We will exemplify it with the proof of equality (2):

x ∈ f ^-1 (A ∩ B) iff f(x) ∈ (A ∩ B) wttw f(x) ∈ A and f(x) ∈ B wttw x ∈ f ^-1(A) and x ∈ f ^-1(B) wttw x ∈ f ^-1 (A) ∩ f ^-1 (B).

Question 3.4.2 Let f : R → R, be a function determined by the formulae f(x) = x ² -5x + 4. Delimit f(R\R⁺ ) and f ^-1 ( {0,4}).