# Least upper bound and positivity axioms

• May 7th 2013, 10:53 AM
Ragnarok
Least upper bound and positivity axioms
I have a rather simple question here which however I feel needs a lot of background info to ask properly. So I apologize in advance!

I'm self-studying with Fitzpatrick's Advanced Calculus and am working on a problem from the very first section, where no calculus has been introduced yet:

Define $S\equiv \{x\mid x\in\mathbb{R},\,x\geq0,\,x^2.
a. Show that $c+1$ is an upper bound for $S$ and therefore, by the completeness axiom, $S$ has a least upper bound that we denote by $b$.
b. Show that if $b^2>c$, then we can choose a suitably small positive number $r$ such that $b-r$ is also an upper bound for $S$, thus contradicting the choice of $b$ as the least upper bound of $S$.
c. Show that if $b^2, then we can choose a suitably small positive number $r$ such that $b+r$ belongs to $S$, thus contradicting the choice of $b$ as an upper bound of $S$.
d. Use parts (b) and (c) and the positivity axioms for $\mathbb{R}$ to conclude that $b^2=c$.

From the hints on that thread I think I have answered parts a-c, however, I have a question on part d. Parts b and c showed that $b^2\leq c$ and $c\leq b^2$, respectively. So it seems obvious that $b^2$ must equal $c$, but I am confused by the reference to the positivity axioms. As stated in the books, they are:

1. If $a$ and $b$ are positive, then $ab$ and $a+b$ are also positive.
2. For a real number $a$, exactly one of the following three alternatives is true: $a$ is positive, $-a$ is positive, $a=0$.

How do these allow us to conclude that $b^2=c$? Sorry for the involved question, it's probably really simple but I have no one to ask.
• May 7th 2013, 11:37 AM
Plato
Re: Least upper bound and positivity axioms
Quote:

Originally Posted by Ragnarok
. As stated in the books, they are:

1. If $a$ and $b$ are positive, then $ab$ and $a+b$ are also positive.
2. For a real number $a$, exactly one of the following three alternatives is true: $a$ is positive, $-a$ is positive, $a=0$.

How do these allow us to conclude that $b^2=c$? Sorry for the involved question, it's probably really simple but I have no one to ask.

$b^2\le c$ means $c-b^2$ is positive or zero.

$c\le b^2$ means $b^2-c=-(c-b^2)$ is positive or zero.

Therefore, the only possible alternative is that $b^2-c=0$ or $b^2=c$.
• May 7th 2013, 07:20 PM
Ragnarok
Re: Least upper bound and positivity axioms
Ah! Got it! Thanks so much again, Plato.
• May 7th 2013, 07:40 PM
Ragnarok
Re: Least upper bound and positivity axioms
Since people here are so helpful, I thought maybe I'd ask to make sure I have parts b and c right. This is my proof for part b; c is similar. I guess where I get confused is what allows us to assume that we can always choose an $r$ such that $2rb-r^2, since the concept of limit has not been developed yet.

Quote:

b. Show that if $b^2>c$, then we can choose a suitably small positive number $r$ such that $b-r$ is also an upper bound for $S$, thus contradicting the choice of $b$ as the least upper bound of $S$.

Suppose $b^2>c$. Then $b^2$ can be written as $c+d$ for some positive number $d$. Then $(b-r)^2 = b^2-2rb+r^2 = c+d-2rb+r^2.$ So $(b-r)^2$ will be larger than $c$ when $d-2rb+r^2>0$, i.e. when $2rb-r^2. So if we choose $r$ such that $2rb-r^2, then we get $(b-r)^2>c$. Then by the definition of $S$, $x for all $x\in S$. So $b-r$ is an upper bound for $S$ which is a smaller upper bound than $b$. Since supposing that $b^2>c$ contradicts our assumption that $b=\text{sup }S$, we must have $b^2\leq c$.