MHB Another supremum and infimum problem

[alexmahone]

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 $\sup S\le\inf T$.

 

[Plato2]

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?$

 

[alexmahone]

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?

 

[Plato2]

I don't know. Why?

Well every $s\in S$ is a lower bound for $T$.
Therefore $s\le\inf(T)$.

 

[alexmahone]

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?)

 

[Plato2]

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.