1. About properties of set R

Can you prove that;
If R has supremum property then R is connected.

Thank you..

2. 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...

3. 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.

4. 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?

5. "R has supremum property" : false assumption
Then any consequence will make the sentence true :P

See Truth table - Wikipedia, the free encyclopedia

By the way, what is the "supremum property" ?

6. If you want i can upload book scan.

7. 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)

9. 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.

10. 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.

11. 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.