Bounded subsets (supremum and infimum) The problem is as follows:
Let $S$ and $T$ be nonempty, bounded subsets of the real numbers. Prove that if $T$ contains $S$, then $\inf T\leq\inf S\leq \sup S\leq\sup T.$
My problem is that I don’t understand why the parent set has the greatest lower bound and the least upper bound. Help. Please and thank you.

We know that $S \subseteq T$, let's see what happens if $\inf S < \inf T$. If this were true, then $\inf T - \inf S > 0$, so there exists an $s \in S$ such that $\inf S \leq s < \inf T$ (to see this: define $\varepsilon = \inf T - \inf S > 0$ which allows us to find some $s \in S$ such that $\inf S \leq s < \inf S + \varepsilon$ and finally conclude $\inf S \leq s < \inf S + \inf T - \inf S = \inf T$).
As we know that $\inf S \leq s < \inf T$, we can state that $s$ is an strict lower bound for $T$, i.e., for all $t \in T$, $s < t$, which is a contradiction since $S \subseteq T$ and thus $s \in T$ ($\implies s<s$). We conclude that $\inf S \geq \inf T$.
By definition we know $\inf S \leq \sup S$ and using the above argument for the supremum we finally have $\inf T \leq \inf S \leq \sup S \leq \sup T$.
I hope it helps :)
