# Simple proof for : if $f(t) \leq g(t)$ for all $t$ then $\sup f(t) \leq \sup g(t)$

Let $t \in (a, \infty )$

if $f(t) \leq g(t)$ for all $t$, then $\sup f(t) \leq \sup g(t)$

Is there a simple proof for this?

Thank you!!

$\sup g$ is an upper bound for $g$. Since $g(t) \ge f(t)$ for all $t$, $\sup g$ is also an upper bound for $f$. Therefore $\sup g\ge \sup f$ since $\sup f$ is the smallest upper bound of $f$.
$$f(t)\leq g(t)\leq sup\ g,\ \forall t\in(a,\infty) \Rightarrow f(t) \leq\ sup\ g$$ meaning $sup\ g$ is an upper quota for $f(t)$. Therefore, by definition of $sup$, $$sup\ f \leq sup\ g$$
Let $A=(a,\infty)$, and let $f(t)\leq g(t)$. With the definition of supremum we have: $$f(t) \leq g(t) \leq \sup\{g(t): t\in A\}$$ So we have $f(t) \leq \sup\{g(t): t\in A\}$. Because $\sup\{g(t): t\in A\}$ is "an" upper bound of $f(t)$ and $\sup\{f(t): t\in A\}$ is the smallest upper bound, and remembering that the supremum is a number. It follows that:
$$f(t) \leq \sup\{f(t): t\in A\} \leq \sup\{g(t): t\in A\}.$$ From which it follows that: $\sup\{f(t): t\in A\} \leq \sup\{g(t): t\in A\}$.