Let $A$ be a non empty set of real numbers and let $f,g$ be functions defined on $A$ such that $f$ is bounded above on $A$ and $g$ is bounded below on $A$ . If $f(x)\le g(x) \;\forall\; x \in A$. Prove that $\sup f(x)\le\sup g(x)$.
And what does $f$ is bounded above on $A$ mean?
To say that f is bounded above on $A$ means that there exists a constant $M$ such that $f(x) \le M$ for all $x \in A$. Similarly for $g$. Let $f(A) = \{f(x):x\in A\}$ and $g(A) = \{g(x) : x \in A\}$. Since $A$ is nonempty and both $f$ and $g$ are bounded above on $A$, $f(A)$ and $g(A)$ are nonempty subsets of $R$ that are bounded above. By completeness, $\sup f(A)$ and $\sup g(A)$ exist. Since $f(x) \le g(x) \le \sup g(A)$ for all $x \in A$, $\sup g(A)$ is an upper bound for $f(A)$. Hence $\sup f(A) \le \sup g(A)$.