1. ## Supremum and infimum

Let $S$ and $T$ be non-empty bounded subsets of $\mathbb{R}$, and suppose that for all $s \in S$ and $t \in T$, we have $s \leq t$.

Prove that supremum of $S \leq$ infimum $T$.

2. Originally Posted by cgiulz
Let $S$ and $T$ be non-empty bounded subsets of $\mathbb{R}$, and suppose that for all $s \in S$ and $t \in T$, we have $s \leq t$.
Prove that supremum of $S \leq$ infimum $T$.
There is no problem knowing that the two exist. Why?
So let $\sigma = \sup (S)\;\& \;\tau = \inf (T)$.
Working for a contradiction, suppose that $\tau < \sigma$.
So $\tau$ is not an upper bound of $S$. Why?
What does that imply about some element of $S$?
3. Since $S$ and $T$ are bounded there must be an infinimum and supremum.
We can see that $\tau$ is not an upper bound of $S$ since it is smaller than the least upper bound.
So clearly there is some element $t \in T$ and $s \in S$ such that $s > t$. So we have our contradiction.
Thus, $Sup S \leq Inf T$.