- #1

Mr Davis 97

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## Homework Statement

Let ##S,T \subseteq \mathbb{F}## be nonempty sets. Assume ##\sup (S)## and ##\sup (T)## both exist in ##\mathbb{F}##. Show that ##\forall a \in \mathbb{F}^+ \cup \{0\}## we have ##\sup(aS) = a \cdot \sup (S)##.

## Homework Equations

## The Attempt at a Solution

First I prove the lemma that if ##B## is the set of upper bounds for ##S## then ##aB## is the set of upper bounds for ##aS##: Let ##x \in aB##. Then ##x = ab## for some upper bound ##b## for ##B##. But ##\forall s \in S~s \le b##, which implies that ##\forall s\in S ~ as \le ab = x##, so ##x## is an upper bound for the set ##aS##

Now we prove the main result. First, we show that ##a \cdot \sup (S)## is an upper bound for ##aS##: Let ##s \in S##. Then ##s \le \sup (S)## by definition. So ##as \le a \cdot \sup (S)##. But ##s## was arbitrary so, ##a \cdot \sup (S)## is an upper bound for ##aS##. Second, we show that ##a \cdot \sup (S)## is less than or equal to any other lower bound of ##aS##. Let ##b## be an upper bound for ##s##. Then ##\sup (S) \le b##, which implies that ##a \cdot \sup(S) \le ab##, but the above lemma shows that ##ab## is an arbitrary upper bound for ##aS##. Hence ##a \cdot \sup (S)## is less than or equal to any other lower bound of ##aS##. So ##\sup(aS) = a \cdot \sup (S)##.