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Another supremum and infimum problem

Alexmahone

Active member
Jan 26, 2012
268
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
Jan 27, 2012
196
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?$
 
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Alexmahone

Active member
Jan 26, 2012
268

Plato

Well-known member
MHB Math Helper
Jan 27, 2012
196

Alexmahone

Active member
Jan 26, 2012
268
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
Jan 27, 2012
196
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.