Prove that supremum of one function of $x$ is less than or equal to another function of $x$

Question

Let $X \subset \mathbb{R}$ and $f,g:X\to\mathbb{R}$ with g bounded above and $f(x)<g(x)$ for all $x \in X$. Prove that $sup(f) \leq sup(g)$.

Also is it necessarily true that $sup(f) < sup(g)$?

Answer 1

Both $f,g$ are bounded above so $\sup f,\sup g$ exist.

Since $f<g\le\sup g,\sup g$ is an upper bound for $f(X)$. Thus $\sup f\le\sup g$.

It is not necessary that $\sup f<\sup g$. Take, for example, $g(x)=k,f(x)=k-1/x,X=\Bbb R^+$. $g$ is bounded above by $k,f<g$ over $\Bbb R^+$ but $\sup g=\sup f=k$.