# least upper bound property

• Apr 29th 2009, 10:42 PM
chaoticmath
least upper bound property
Hi can I get some help with the following theorem I need to prove:

Let A be a subset of the ordered set X. If A has an upper bound then A has a least upper bound..

Thanks!
• Apr 30th 2009, 04:14 AM
Showcase_22
Quote:

If A has an upper bound then A has a least upper bound.
I think it can be done this way:

From the question we know that A has an upper bound. Let's suppose A has two upper bounds $\displaystyle \alpha$ and $\displaystyle \beta$ with $\displaystyle \alpha \neq \beta$.

$\displaystyle \alpha \neq \beta \Rightarrow \ \alpha < \beta$ ($\displaystyle \beta < \alpha$ is also valid but i've decided to choose $\displaystyle \alpha < \beta$).

Hence A has a least upper bound which is $\displaystyle \alpha$.
• Apr 30th 2009, 04:26 AM
Plato
What if $\displaystyle X = \mathbb{Q}\;\& \;A = \left\{ {x \in \mathbb{Q}:x^2 < 2} \right\}$, $\displaystyle \mathbb{Q}$ is the set of rationals.
Is $\displaystyle A$ bounded above by 2?
But can $\displaystyle A$ have a least upper bound? No!
So the statement if false without some more given conditions.
• Apr 30th 2009, 04:28 AM
Showcase_22
but isn't the least upper bound $\displaystyle \sqrt{2}$, it's just that $\displaystyle \sqrt{2}$ isn't in the set?
• Apr 30th 2009, 04:34 AM
Plato
Quote:

Originally Posted by Showcase_22
but isn't the least upper bound $\displaystyle \sqrt{2}$, it's just that $\displaystyle \sqrt{2}$ isn't in the set?

That is the whole point! The LUB must be in the space you are dealing with.
In that example, all you have is the set of rational numbers.
You do not have $\displaystyle \sqrt 2$ to even consider.