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Set of least upper bounds multiplied by a constant

  • #1
1,456
44

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)##.
 

Answers and Replies

  • #2
12,633
9,148

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 "of" or "##\in##"
##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)##.
Instead of having this abstract field, you should have better said something about the order required for ##\mathbb{F}##, as it cannot be arbitrary for this reason. Also you use ##a\leq b \Longrightarrow ca\leq cb ## for ##c\geq 0## which is again a condition the order must have. I'm not quite sure whether you need this Lemma first as I think you could directly show the inequalities. That's the difficulty with such "obvious" results, to figure out what has to be shown at all. It looks o.k. so far, even if perhaps a bit too long at the wrong places.
 

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