Real analysis Supremum and Infinum

Question

Let $S$ and $T$ be subsets of $\Bbb{R}$ such that $s < t$ for each $s ∈ S$ and each $t ∈ T$. Prove carefully that $\sup S ≤ \inf T$.

This question has been posted a few times, but I don't think the answers are formal enough

I start by using the definition for supremum and infinum, $\sup(S)= a$ and $\inf(T)= b$ and I know that $a> s$ and $b< t$ for all $s$ and $t$. How do I continue? Do I prove it directly starting from $s< t$ or will it be easier to use proof by contradiction?

Answer 1

Hint:

If s < t for all elements from S and T, then $S \cap T$ is empty. Now consider the number line and finish the proof.

Answer 2

Let $a=\sup S$ and $b=\inf T$, as you suggested. When know that $s<t$ for all $s\in S,t\in T$. So, fix $t_{0}\in T$. Then, since $s<t_{0}$ for all $s\in S$, it must be that $a\leq t_{0}$; otherwise, by the definition of $\sup$, there would be an $s\in S$ such that $a\geq s>t$, which is impossible. Now, $t_{0}$ was chosen arbitrarily, so $a\leq t$ must hold for all $t\in T$. Hence, reasoning as above, it follows that $a\leq b$.

Answer 3

Let $s\in S$ and $t\in T$ be arbitrary. By hypothesis, $s<t ≤\sup T$. Thus $\sup T$ is begger than or equal to all $s\in S$, so by definition of sup (the SMALESST upper bound), we must have $\sup S≤\sup T$.