# Least Upper Bound Proof

• December 2nd 2012, 12:51 PM
JBrandon
Least Upper Bound Proof
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.

definition i have to use:

The number q is an UPPER BOUND for the set A if x<=q for all x in A.
The LEAST UPPER BOUND of the nonempty set A is L if L is an upper bound for A and if r is a number less than L, then there is a member a in A so that a > r.

what i have so far.

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. my initial thought was to just use the same trick as above but this feels flawed for some reason.

since L is the least upper bound of A, then r < a <=L where r is a number less than L.

since r < a <=L where r is a number less than L, and p is positive, then rp < pa <= pL.

since rp < pa <= pL, and rp is a number less than pL, then pL is the least upper bound of pA.
• December 2nd 2012, 01:06 PM
Plato
Re: Least Upper Bound Proof
Quote:

Originally Posted by 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 $\lambda=\text{LUB}(pA)$. Suppose that $\lambda. (We know that $\lambda\le pL$ WHY?)

It follows that $\frac{\lambda}{p} so $(\exists a'\in A)\left[\frac{\lambda}{p} WHY?

• December 2nd 2012, 01:55 PM
JBrandon
Re: Least Upper Bound Proof
Quote:

Originally Posted by Plato
Let $\lambda=\text{LUB}(pA)$. Suppose that $\lambda. (We know that $\lambda\le pL$ 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.

Quote:

Originally Posted by Plato
It follows that $\frac{\lambda}{p} so $(\exists a'\in A)\left[\frac{\lambda}{p} WHY?

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

Quote:

Originally Posted by Plato

not to sure on this one. is it because $\frac{\lambda}{p}$ is supposed to equal L but by the proof is less than L?
• December 2nd 2012, 01:59 PM
Plato
Re: Least Upper Bound Proof
Quote:

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

Well that implies that $\lambda. Can that be?
• December 2nd 2012, 04:52 PM
JBrandon
Re: Least Upper Bound Proof
so the full proof should be.

Let $A \neq \phi$, $L=\text{LUB}(A)$, $a \in A$, $pa \in pA$, and p be a positive number.
Since $A \neq \phi$, and $L=\text{LUB}(A)$, then $a \le L$.
Since $a \le L$, then $pa \le pL$.
therefor $pL=\text{UB}(pA)$.
Assume $\lambda=\text{LUB}(pA)$
since $\lambda=\text{LUB}(pA)$, and $pL=\text{UB}(pA)$, then $\lambda \le pL$.
since $\lambda \le pL$, then $\frac{\lambda}{p} < L$
since $\frac{\lambda}{p} < L$, and $L=\text{LUB}(A)$, then $\frac{\lambda}{p} < a \le L$
since $\frac{\lambda}{p} < a \le L$, then $\lambda < pa \le pL$. a contradiction as $pa \le \lambda$ must be true for $\lambda=\text{LUB}(pA)$
therefor $pL < \lambda$ and $pL = \text{LUB}(pA)$
• December 2nd 2012, 05:42 PM
Plato
Re: Least Upper Bound Proof
Quote:

Originally Posted by JBrandon
so the full proof should be.
Let $A \neq \phi$, $L=\text{LUB}(A)$, $a \in A$, $pa \in pA$, and p be a positive number.
Since $A \neq \phi$, and $L=\text{LUB}(A)$, then $a \le L$.
Since $a \le L$, then $pa \le pL$.
therefor $pL=\text{UB}(pA)$.
Assume $\lambda=\text{LUB}(pA)$
since $\lambda=\text{LUB}(pA)$, and $pL=\text{UB}(pA)$, then $\lambda \le pL$.
since $\lambda \le pL$, then $\frac{\lambda}{p} < L$
since $\frac{\lambda}{p} < L$, and $L=\text{LUB}(A)$, then $\frac{\lambda}{p} < a \le L$
since $\frac{\lambda}{p} < a \le L$, then $\lambda < pa \le pL$. a contradiction as $pa \le \lambda$ must be true for $\lambda=\text{LUB}(pA)$
therefor $pL < \lambda$ and $pL = \text{LUB}(pA)$

Honestly, I don't know how to answer you questions.
I taught analysis for more than 35 years.
I drilled into students that if $\lambda=\text{LUB}(A)$ then for any $c>0$ then it must be the case that $(\exists t\in A)[\lambda-c.

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

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