Can you prove that;

If R has supremum property then R is connected.

Thank you..

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- Jul 9th 2009, 07:31 AMalper82About properties of set R
Can you prove that;

If R has supremum property then R is connected.

Thank you.. - Jul 9th 2009, 07:34 AMProve It
- Jul 9th 2009, 07:42 AMalper82
yes i am talking about real numbers. I think so but i saw this thorem in a book. But maybe they are talking about any subset (interval) of R.

- Jul 9th 2009, 07:44 AMProve It
Well let me put it this way...

The set of real numbers contains infinitely many terms, each one greater than the last.

How can there be a supremum? - Jul 9th 2009, 07:47 AMMoo
"R has supremum property" : false assumption

Then any consequence will make the sentence true :P

See Truth table - Wikipedia, the free encyclopedia :D

By the way, what is the "supremum property" ? - Jul 9th 2009, 07:52 AMalper82
If you want i can upload book scan.

- Jul 9th 2009, 08:02 AMalper82
This is theorem..

http://img36.imageshack.us/img36/2812/pictureqlj.jpg

and this is proof..

Imageshack - picture001c

sup özelliği = sup property

bağlantılıdır = connected

(this is a p->q, ~q->~p proof) - Jul 9th 2009, 10:53 AMalper82
Another proof please?

- Jul 9th 2009, 11:01 AMpomp
Surely the set $\displaystyle \left\{0,\frac{1}{2},\frac{3}{4}, \frac{7}{8}, \frac{15}{16}, \ldots \right\} $ has infinitely many terms, each one greater than the last. Clearly has supremum $\displaystyle 1$ though.

When I learned about supremums we were taught the convention that an empty set has supremum $\displaystyle - \infty$ and a set unbounded above has supremum $\displaystyle \infty$, so the set of real numbers has supremum $\displaystyle \infty $, this is known as an affine extension.

Regardless, this question still doesn't make much sense. - Jul 9th 2009, 11:16 AMpomp
Taking the supremum property to mean:

Every bounded subset of $\displaystyle \mathbb{R}$ has a supremum in $\displaystyle \mathbb{R}$,

the proof goes as follows.

Let a subset $\displaystyle U$ of $\displaystyle \mathbb{R}$ be both open and closed.

Take $\displaystyle a \in U$ and using our supremum property, take b as the supremum of real numbers such that $\displaystyle [a,b) \subset U$

If $\displaystyle b<\infty$, then b is a limit point of U and since U is closed, $\displaystyle b \in U.$

Using the assumption that $\displaystyle U$ is also open we also must have that

$\displaystyle \exists$ $\displaystyle \epsilon > 0$ such that $\displaystyle (b-\epsilon, b+\epsilon) \subset U$

Hence $\displaystyle [a, b + \epsilon) \subset U$ which contradicts b being a supremum. Therefore $\displaystyle b = \infty.$

Repeating this arguement for an infimum we have that $\displaystyle U = \mathbb{R}$ or $\displaystyle U = \oslash $

Therefore $\displaystyle \mathbb{R}$ is connected.

Hope this cleared some things up

Pomp. - Jul 9th 2009, 11:37 AMSampras