Question

In: Advanced Math

Problem 17.5. Consider the function χ(0,1) : R → R (this is the characteristic of Definition...

Problem 17.5. Consider the function χ(0,1) : R → R (this is the characteristic of Definition 14.4). Find: (a) χ(0,1)((0,1)); (b) χ(0,1)((−1,3)); (c) and (in general) χ(0,1)((a,b)), where a,b ∈ R and a < b; prove that the set you found is correct; (d) χ−1 (0,1) ((−2,−1)); (e) χ−1 (0,1) ((0,2)); (f) and (in general) χ−1 (0,1) ((a,b)), where a,b ∈ R and a < b; prove that the set you found is correct,

Definition 14.4 (for Problems 14.6 through 14.9). Let X be a nonempty set and let
A be a subset of X. The characteristic function or indicator function of the set A
in X is

χA : X → {0,1} defined by χA(x) = 1 if x ∈ A
0 if x ∈ X\A

Solutions

Expert Solution

From definition, is given by .

(a). . This follows trivially from the above stated definition.

(b). as for any x in , so, and , combining them we get .

(c). Case 1.:- . Then, for any we have . So, .

Case 2.:- . Here, 0 is in (a,b) and and , so . Hence, .

Case 3.:- . Then, similarly as case 2, .

Case 4:- . Here, (a,b) is a subset of (0,1) so, .

Case 5:- . Then, (a,1) is a subset of (0,1), so and since , . Therefore, .

Case 6:- . Here , so .

Now observe that, and .

(d). As 0 and 1 are both not in (-2,-1), we get where denotes the empty set.

(e). As we get .

(f). Case 1.- . Here, so we get .

Case 2.- . Here, so we get .

Case 3.- . Here, , so we get .

Case 4.- . Here, , so we get .

Case 5.- . Here, so we get .

Case 6.- . Here, so we get .

Colby Messinger answered 2 years ago
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