Another supremum and infimum problem

Alexmahone

Active member
Let $S$ and $T$ be non-empty subsets of $\mathbb{R}$, and suppose that for all $s\in S$ and $t\in T$, we have $s\le t$. Prove that $\sup S\le\inf T$.

Plato

Well-known member
MHB Math Helper
Let $S$ and $T$ be non-empty subsets of $\mathbb{R}$, and suppose that for all $s\in S$ and $t\in T$, we have $s\le t$. Prove that $\sup S\le\inf T$.
Clearly $\inf(T)$ & $\sup(S)$ exist. WHY?

$\forall s\in S$ we know $s\le\inf(T)~.$ WHY?$Last edited: Alexmahone Active member Clearly$ \inf(T)$&$\sup(S) $exist. WHY? Because$S$and$T$are non-empty subsets of$\mathbb{R}$.$\forall s\in S$we know$s\le\inf(T)~.$WHY? I don't know. Why? Plato Well-known member MHB Math Helper I don't know. Why? Well every$s\in S$is a lower bound for$T$. Therefore$s\le\inf(T)$. Alexmahone Active member Well every$s\in S$is a lower bound for$T$. Therefore$s\le\inf(T)$. So if$\sup S>\inf T$, there must be an$s\in S$such that$s>\inf T$. (Otherwise,$\inf T$is a smaller upper bound for$S$than$\sup S$.) So we get a contradiction. (Is that correct?) Plato Well-known member MHB Math Helper So if$\sup S>\inf T$, there must be an$s\in S$such that$s>\inf T$. (Otherwise,$\inf T$is a smaller upper bound for$S$than$\sup S\$.) So we get a contradiction. (Is that correct?)
Yes. It is correct.