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Math Help - About properties of set R

  1. #1
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    About properties of set R

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

    Thank you..
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  2. #2
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    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...
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  3. #3
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    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.
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  4. #4
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    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?
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  5. #5
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    "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" ?
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  6. #6
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    If you want i can upload book scan.
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  7. #7
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    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)
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  8. #8
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    Another proof please?
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  9. #9
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    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.
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  10. #10
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    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.
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  11. #11
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    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.
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