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#### Alexmahone

##### Active member

- Jan 26, 2012

- 268

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- #1

- Jan 26, 2012

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Clearly $ \inf(T)$ & $\sup(S) $ exist. WHY?

$\forall s\in S$ we know $s\le\inf(T)~.$ WHY?$

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- Jan 26, 2012

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Because $S$ and $T$ are non-empty subsets of $\mathbb{R}$.Clearly $ \inf(T)$ & $\sup(S) $ exist. WHY?

I don't know. Why?$\forall s\in S$ we know $s\le\inf(T)~.$ WHY?

Well every $s\in S$ is a lower bound for $T$.I don't know. Why?

Therefore $s\le\inf(T)$.

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- Jan 26, 2012

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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?)Well every $s\in S$ is a lower bound for $T$.

Therefore $s\le\inf(T)$.

Yes. It is correct.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?)