# About properties of set R

• Jul 9th 2009, 08:31 AM
alper82
About properties of set R
Can you prove that;
If R has supremum property then R is connected.

Thank you..
• Jul 9th 2009, 08:34 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 $\mathbf{R}$, as in the set of all real numbers?

The set of real numbers does not have a supremum...
• Jul 9th 2009, 08:42 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, 08:44 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, 08:47 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, 08:52 AM
alper82
If you want i can upload book scan.
• Jul 9th 2009, 09:02 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, 11:53 AM
alper82
• Jul 9th 2009, 12:01 PM
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 $\left\{0,\frac{1}{2},\frac{3}{4}, \frac{7}{8}, \frac{15}{16}, \ldots \right\}$ 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.
• Jul 9th 2009, 12:16 PM
pomp
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.
• Jul 9th 2009, 12:37 PM
Sampras
Quote:

Originally Posted by pomp
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.