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!
I think it can be done this way:If A has an upper bound then A has a least upper bound.
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 $\displaystyle \alpha$.
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.