Question

Let X be a nonempty set and let f: X-> R have bounded ranges in r. if a element R. show that : sup {a + f(x): x element X} = a + sup {f(x): x element X} .Show that we also have inf {a+f(x) : x element X} = a+ inf {f(x): x element X}

Answer 1

Answer on question 34185 – Math – Real Analysis

Let $X$ be a nonempty set and let $f: X \text{-} \& \text{gt}$; $R$ have bounded ranges in $r$. If a element $R$ show that: $\sup \{a + f(x): x \text{ element } X\} = a + \sup \{f(x): x \text{ element } X\}$. Show that we also have $\inf \{a + f(x): x \text{ element } X\} = a + \inf \{f(x): x \text{ element } X\}$.

Solution

Let $y = \sup \{a + f(x): x \in X\}$. This is mean that $y$ is a smaller real number such that $\forall x \in X: a + f(x) \leq y$. From the last inequality we get

$$f(x) \leq y - a, \forall x \in X$$

Therefore $y$-a is the supremum for $f(x)$, we get it from the definition. Really

$$y = a + y - a.$$

Similarly for the infimum. If $z = \inf \{a + f(x): x \text{ element } X\}$ then $z$ is the larger real number such that $\forall x \in X: a + f(x) \geq z$. From the last inequality we get

$$f(x) \geq z - a, \forall x \in X$$

$z$-a is the larger real number for which the last inequality satisfies. From the definition of infimum we obtain that $z$-a is the infimum of $f(x)$. Substituting these into equality we get the identity

$$z = a + z - a.$$

QED.

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