Question

let x be a nonempty set and let f: X->R have bounded range in R. if a element R , show that
sup(a+f(x): x element X)= a + sup (f(x): x element X)
and
inf(a+f(x): x element X)= a + inf (f(x): x element X)

Answer 1

Conditions

let x be a nonempty set and let f: X->R have bounded range in R. if a element R, show that

$\sup(a + f(x): x \text{ element } X) = a + \sup(f(x): x \text{ element } X)$

and

$\inf(a + f(x): x \text{ element } X) = a + \inf(f(x): x \text{ element } X)$

Solution

$$X \neq \emptyset, f: X \to R, a \in R$$

Consider

$$s = \sup (\{a + f (x) \}) \leftrightarrow s \in S _ {x} \forall y \in S _ {x}: s \leq y, \text{ where } S _ {x} = \{y \in M | \forall x \in X: x \leq y \}$$ $$\forall x \in X, a + f (x) \leq s \leq y$$ $$\forall x \in X, f (x) \leq s - a$$ $$s - a \text{ is a supremum for } \{f (x), x \in X \}$$

Consider

$$a + \sup (\{f (x) \}) = s - a + a = s, \text{ which is } \sup (\{a + f (x) \})$$

The same way is how to proof that

$$\inf (\{a + f (x) \}) = a + \inf (\{f (x) \})$$ $$\inf (\{f (x) \}) = i, \forall x \in X \text{ } x \geq i, \forall x \in X \text{ } x + a \geq i + a$$ $$\text{So, } \inf (\{a + f (x) \}) = i + a = a + \inf (\{f (x) \})$$