Supremum and infimum of a bounded real valued function

Question

I was reading a real analysis book online about supremum and infimum. When I came across this.

But it did not have any proof. I don’t see how it is «clear» Would anyone mind proving it or explaining?

Answer 1

As f is defined on an non-empyset, $f(D)$ also is a nonempty set, therefore both $\inf f$ and $\sup f$ exist, and the inequalities hold. If they didn't, we'd have that there is an element $x_0 \in D$ such that $f(x_0)< \inf f$ or $f(x_0)>\sup f$, which contradicts the definition of $\inf$ and $\sup$

Answer 2

You will always have, for $x \in D$ that $\inf f = \inf_{y \in D} f(y) \le f(x) \le \sup_{y \in D} f(y) = \sup f$.

This is always true with the convention that $\inf_{y \in D} f(y) = -\infty$ if $f$ is unbounded below on $D$ and similarly for $\sup$.