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!

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- Apr 29th 2009, 10:42 PMchaoticmathleast 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 AMShowcase_22Quote:

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 $\displaystyle \alpha$. - Apr 30th 2009, 04:26 AMPlato
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 AMShowcase_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 AMPlato