Can you prove that;

If R has supremum property then R is connected.

Thank you..

Printable View

- Jul 9th 2009, 08:31 AMalper82About properties of set R
Can you prove that;

If R has supremum property then R is connected.

Thank you.. - Jul 9th 2009, 08:34 AMProve It
- Jul 9th 2009, 08:42 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, 08:44 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, 08:47 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, 08:52 AMalper82
If you want i can upload book scan.

- Jul 9th 2009, 09:02 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, 11:53 AMalper82
Another proof please?

- Jul 9th 2009, 12:01 PMpomp
Surely the set has infinitely many terms, each one greater than the last. Clearly has supremum though.

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

Regardless, this question still doesn't make much sense. - Jul 9th 2009, 12:16 PMpomp
Taking the supremum property to mean:

Every bounded subset of has a supremum in ,

the proof goes as follows.

Let a subset of be both open and closed.

Take and using our supremum property, take b as the supremum of real numbers such that

If , then b is a limit point of U and since U is closed,

Using the assumption that is also open we also must have that

such that

Hence which contradicts b being a supremum. Therefore

Repeating this arguement for an infimum we have that or

Therefore is connected.

Hope this cleared some things up

Pomp. - Jul 9th 2009, 12:37 PMSampras