Re: Least Upper Bound Proof

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**JBrandon** IF A is a non-empty bounded set of real numbers with least upper bound L and p is a positive real number, then the set pA has least upper bound pL.

Since A is non empty and the least upper bound of A is L, then a<=L for all a in A.

Since a<=L for all a in A and p is positive, then pa<=pL.

therefor pL is a upper bound of pA.

now im lost on the second part of this problem where i need to prove that pL is the least upper bound of pA.

Let $\displaystyle \lambda=\text{LUB}(pA) $. Suppose that $\displaystyle \lambda<pL$. (We know that $\displaystyle \lambda\le pL$ WHY?)

It follows that $\displaystyle \frac{\lambda}{p}<L$ so $\displaystyle (\exists a'\in A)\left[\frac{\lambda}{p}<a'\le L\right]$ WHY?

How is that a contradiction?

Re: Least Upper Bound Proof

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**Plato** Let $\displaystyle \lambda=\text{LUB}(pA) $. Suppose that $\displaystyle \lambda<pL$. (We know that $\displaystyle \lambda\le pL$ WHY?)

since i know pL is a upper bound of pA the least upper bound of pA must be less than or equal to pL.

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**Plato** It follows that $\displaystyle \frac{\lambda}{p}<L$ so $\displaystyle (\exists a'\in A)\left[\frac{\lambda}{p}<a'\le L\right]$ WHY?

since i know L is the least upper bound of A, then $\displaystyle \frac{\lambda}{p}<a'$

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**Plato** How is that a contradiction?

not to sure on this one. is it because $\displaystyle \frac{\lambda}{p}$ is supposed to equal L but by the proof is less than L?

Re: Least Upper Bound Proof

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**JBrandon** not to sure on this one. is it because $\displaystyle \frac{\lambda}{p}$ is supposed to equal L but by the proof is less than L?

Well that implies that $\displaystyle \lambda<pa'$. Can that be?

Re: Least Upper Bound Proof

so the full proof should be.

Let $\displaystyle A \neq \phi$, $\displaystyle L=\text{LUB}(A)$, $\displaystyle a \in A$, $\displaystyle pa \in pA$, and p be a positive number.

Since$\displaystyle A \neq \phi$, and $\displaystyle L=\text{LUB}(A)$, then $\displaystyle a \le L$.

Since $\displaystyle a \le L$, then $\displaystyle pa \le pL$.

therefor $\displaystyle pL=\text{UB}(pA)$.

Assume $\displaystyle \lambda=\text{LUB}(pA)$

since $\displaystyle \lambda=\text{LUB}(pA)$, and $\displaystyle pL=\text{UB}(pA)$, then $\displaystyle \lambda \le pL$.

since $\displaystyle \lambda \le pL$, then $\displaystyle \frac{\lambda}{p} < L$

since $\displaystyle \frac{\lambda}{p} < L$, and $\displaystyle L=\text{LUB}(A)$, then $\displaystyle \frac{\lambda}{p} < a \le L$

since $\displaystyle \frac{\lambda}{p} < a \le L$, then $\displaystyle \lambda < pa \le pL$. a contradiction as $\displaystyle pa \le \lambda$ must be true for $\displaystyle \lambda=\text{LUB}(pA)$

therefor $\displaystyle pL < \lambda$ and $\displaystyle pL = \text{LUB}(pA)$

Re: Least Upper Bound Proof

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**JBrandon** so the full proof should be.

Let $\displaystyle A \neq \phi$, $\displaystyle L=\text{LUB}(A)$, $\displaystyle a \in A$, $\displaystyle pa \in pA$, and p be a positive number.

Since$\displaystyle A \neq \phi$, and $\displaystyle L=\text{LUB}(A)$, then $\displaystyle a \le L$.

Since $\displaystyle a \le L$, then $\displaystyle pa \le pL$.

therefor $\displaystyle pL=\text{UB}(pA)$.

Assume $\displaystyle \lambda=\text{LUB}(pA)$

since $\displaystyle \lambda=\text{LUB}(pA)$, and $\displaystyle pL=\text{UB}(pA)$, then $\displaystyle \lambda \le pL$.

since $\displaystyle \lambda \le pL$, then $\displaystyle \frac{\lambda}{p} < L$

since $\displaystyle \frac{\lambda}{p} < L$, and $\displaystyle L=\text{LUB}(A)$, then $\displaystyle \frac{\lambda}{p} < a \le L$

since $\displaystyle \frac{\lambda}{p} < a \le L$, then $\displaystyle \lambda < pa \le pL$. a contradiction as $\displaystyle pa \le \lambda$ must be true for $\displaystyle \lambda=\text{LUB}(pA)$

therefor $\displaystyle pL < \lambda$ and $\displaystyle pL = \text{LUB}(pA)$

Honestly, I don't know how to answer you questions.

I taught analysis for more than 35 years.

I drilled into students that if $\displaystyle \lambda=\text{LUB}(A)$ then for any $\displaystyle c>0$ then it must be the case that $\displaystyle (\exists t\in A)[\lambda-c<t\le \lambda]$.

Oddly enough, that turns out to be the most important idea in analysis.

So if $\displaystyle \lambda<pa'$ because $\displaystyle pa'\in pA$ that must be a contradiction because $\displaystyle pa'\le\lambda~.$