# Set of least upper bounds multiplied by a constant

• Mr Davis 97
In summary, the statement shows that if we have two nonempty sets with supremums in a specific ordered field, then for any positive or zero element in that field, the supremum of the product of that element and the first set is equal to the product of that element and the supremum of the first set. This is proven by first establishing a lemma and then using it to prove the main result.
Mr Davis 97

## 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)##.

## 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)##.

Mr Davis 97 said:

## 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)##.

## 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.

## 1. What is a set of least upper bounds?

A set of least upper bounds is a mathematical concept that refers to the smallest possible upper bound for a given set of numbers. It is also known as the supremum or least upper bound.

## 2. How is a set of least upper bounds determined?

A set of least upper bounds is determined by finding the smallest number that is greater than or equal to all the numbers in a given set. This number is known as the supremum or least upper bound.

## 3. What does it mean to multiply a set of least upper bounds by a constant?

Multiplying a set of least upper bounds by a constant means to multiply the supremum or least upper bound of a set by a given number. This results in a new set of numbers that are all multiplied by the same constant.

## 4. How does multiplying a set of least upper bounds by a constant affect the set?

Multiplying a set of least upper bounds by a constant will result in a new set with a new least upper bound. The new least upper bound will be the original least upper bound multiplied by the constant. The rest of the numbers in the set will also be multiplied by the constant.

## 5. What are some real-world applications of using a set of least upper bounds multiplied by a constant?

One example of a real-world application of using a set of least upper bounds multiplied by a constant is in financial analysis, where a company's financial data may be multiplied by a constant to adjust for inflation or other factors. This concept is also used in mathematical modeling and optimization problems.

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