About properties of set R

alper82 · #1 $Old$ July 9th, 2009, 09:31 AM

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

Thank you..

Prove It · #2 $Old$ July 9th, 2009, 09:34 AM

Quote:

Originally Posted by alper82 $View Post$

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

Thank you..

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

The set of real numbers does not have a supremum...

alper82 · #3 $Old$ July 9th, 2009, 09:42 AM

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.

Prove It · #4 $Old$ July 9th, 2009, 09:44 AM

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?

Moo · #5 $Old$ July 9th, 2009, 09:47 AM

"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" ?

alper82 · #6 $Old$ July 9th, 2009, 09:52 AM

If you want i can upload book scan.

alper82 · #7 $Old$ July 9th, 2009, 10:02 AM

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)

alper82 · #8 $Old$ July 9th, 2009, 12:53 PM

Another proof please?

pomp · #9 $Old$ July 9th, 2009, 01:01 PM

Quote:

Originally Posted by Prove It $View Post$

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

How can there be a supremum?

Surely the set $\left\{0,\frac{1}{2},\frac{3}{4}, \frac{7}{8}, \frac{15}{16}, \ldots \right\}$ has infinitely many terms, each one greater than the last. Clearly has supremum $1$ though.

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

Regardless, this question still doesn't make much sense.

pomp · #10 $Old$ July 9th, 2009, 01:16 PM

Taking the supremum property to mean:

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

the proof goes as follows.

Let a subset $U$ of $\mathbb{R}$ be both open and closed.
Take $a \in U$ and using our supremum property, take b as the supremum of real numbers such that $[a,b) \subset U$

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

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

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

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

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

Therefore $\mathbb{R}$ is connected.

Hope this cleared some things up
Pomp.

Sampras · #11 $Old$ July 9th, 2009, 01:37 PM

Quote:

Originally Posted by pomp $View Post$

Taking the supremum property to mean:

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

the proof goes as follows.

Let a subset $U$ of $\mathbb{R}$ be both open and closed.
Take $a \in U$ and using our supremum property, take b as the supremum of real numbers such that $[a,b) \subset U$

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

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

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

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

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

Therefore $\mathbb{R}$ is connected.

Hope this cleared some things up
Pomp.

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