# Proofs involving the Least Upper and Greatest Lower Bound Properties

Let $S \subset T \subset \Bbb{R}$, then

• if $T$ is bounded from above, then $\sup S \le \sup T$

• if $T$ is bounded from below, then $\inf T \le \inf S$

• if $T$ is bounded, then $\inf T \le \inf S \le \sup S \le \sup T$.

Thanks for the support!

We are of course assuming that $S$ and hence $T$ is nonempty. Since $\sup T\ge x$ for all $x\in T$, we must have $\sup T\ge x$ for all $x\in S$. So, $\sup T$ an upper bound of $S$, hence $\sup S\le\sup T$, since $\sup S$ is the smallest upper bound of $S$.