1. ## Supremum and infimum

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

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

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