Prove that $\sup S \leq \inf T$

Question

Could you help me here?

Let $S$ and $T$ be nonempty subsets of $\mathbb{R}$ and suppose that for all $s \in S $ and $t \in T$ we have $s \leqslant t$. Prove that $\sup S \leqslant \inf T $

PS: My idea was to try 2 cases: when $S \subset T$ and $ S \cap T = \emptyset $. Am I right?

I was able to prove that $t \geq \sup S$ but the final part I couldn't.

Any help?

Answer 1

No need to split into cases. You can proceed directly by the definition here. For any $t \in T$ it follows immediately that $t$ is an upper-bound on $S$. By definition, $\sup S \leq t$. But, now we have $\sup S$ is a lower bound on $T$ since $t$ was arbitrary, so $\sup S \leq \inf T$.

Answer 2

Let $(s_n)_{n=1}^\infty\subset S$ and $(t_n)_{n=1}^\infty\subset T$ be such that $$\lim_{n\to\infty} s_n =\sup S\quad \text{and}\quad\lim_{n\to\infty} t_n =\inf T.$$ Since we have $s_n\leq t_n$ for all $n$, the desired inequality follows.

Answer 3

You can also prove it easily by contradiction.

• $\color{blue}{\text{Assume}}$ $\inf T < \sup S$.
• Because of the supremum property, there is an $s_0 \in S$ such that $\inf T < s_0 \leq \sup S$.
• Because of the infimum property, there is now a $t_0 \in T$ such that $\inf T \leq t_0 \color{blue}{<} s_0 \Rightarrow \color{blue}{\text{Contradiction!}}$ (to $s\leq t$ for all $s\in S$ and $t\in T$)