# About properties of set R

• Jul 9th 2009, 07:31 AM
alper82
Can you prove that;
If R has supremum property then R is connected.

Thank you..
• Jul 9th 2009, 07:34 AM
Prove It
Quote:

Originally Posted by alper82
Can you prove that;
If R has supremum property then R is connected.

Thank you..

Are you talking about the set $\displaystyle \mathbf{R}$, as in the set of all real numbers?

The set of real numbers does not have a supremum...
• Jul 9th 2009, 07:42 AM
alper82
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 AM
Prove 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 AM
Moo
"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 AM
alper82
If you want i can upload book scan.
• Jul 9th 2009, 08:02 AM
alper82
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 AM
alper82
• Jul 9th 2009, 11:01 AM
pomp
Quote:

Originally Posted by Prove It
The set of real numbers contains infinitely many terms, each one greater than the last.

How can there be a supremum?

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 AM
pomp
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 AM
Sampras
Quote:

Originally Posted by pomp
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.

Every bounded subset of an ordered set $\displaystyle S$ has a supremum in $\displaystyle S$ would suffice.