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Math Help - Complement of nonempty open set

  1. #1
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    Complement of nonempty open set

    I've tried to think this, help would be appreciated to get me going a gain.

    If we have two nonempty open set A,B \subset {R}^{n} so that  A \cap B = \emptyset (two separate sets). How could I show that complement of union A and B is also nonempty, \complement (A \cup B)  \neq \emptyset.

    I'm thinking something like, because the complement is {R}^{n} \setminus (A \cup B) so because its R^n there must be some points that belong to that complement. But basicly I think that don't work (just to say because it's R^n) because I have show that it (complement of the union is nonempty) follows from the notion that A,B are open nonempty sets.

    Any ideas and help? Thanks!
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  2. #2
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    Re: Complement of nonempty open set

    R^n is connected so it can't be the union of two components. Thus R^n\(AUB) is non empty.
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    Re: Complement of nonempty open set

    R^1 connected is quite easy to show. How about R^n?
    Last edited by mrjaytski; November 21st 2012 at 02:53 AM.
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    Re: Complement of nonempty open set

    Quote Originally Posted by mrjaytski View Post
    R^1 connected is quite easy to show. How about R^n?
    It is also easy.
    Start with \mathbb{R}^2. Then expand.
    Post what you come up with.
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    Re: Complement of nonempty open set

    Is there, by the way, any other way to show this complement to be non empty? Without using connectivity.
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    Re: Complement of nonempty open set

    Quote Originally Posted by mrjaytski View Post
    Is there, by the way, any other way to show this complement to be non empty? Without using connectivity.
    Why do you want to do that?

    Quote Originally Posted by mrjaytski View Post
    If we have two nonempty open set A,B \subset {R}^{n} so that  A \cap B = \emptyset (two separate sets). How could I show that complement of union A and B is also nonempty, \complement (A \cup B)  \neq \emptyset.

    The OP is an equivalent way of asking for a proof that \mathbb{R}^n is connected. In fact, in Moore spaces that is the very definition of connectivity.

    Go back to my first reply. Can you think of \mathbb{R}^1 as a subset of \mathbb{R}^2~?
    If \mathbb{R}^2 is not connected the do we have a contradiction?
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    Member ModusPonens's Avatar
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    Re: Complement of nonempty open set

    If a space is path connected, it is connected. Clearly R^n is path connected.
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    Re: Complement of nonempty open set

    if the complement is empty, then it means as "ModusPonens" stated, R^n is union of two distinct sets A & B.
    consider a point P on boundary of A. the boundary of A is not belongs to A (cause it is open). but if the complement is empty, this boundary must belong to B. the point P can not belong to interior of B (because of in each neighbor of P, at least one point of A exist), since it must belong to boundary of B. this is contradiction, since it means that P is not belong to B.
    So at least P is not belong to A & B and since belongs to complement.
    i'm not a professional math expert, but p.s somone evaluate this prove.
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    Re: Complement of nonempty open set

    Quote Originally Posted by ansarey View Post
    if the complement is empty, then it means as "ModusPonens" stated, R^n is union of two distinct sets A & B.
    consider a point P on boundary of A.
    How do you know that the boundary of A is not empty?
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    Re: Complement of nonempty open set

    you are right.
    i suppose another proof.
    the complement of open set is closed. so union of A and its complement equal to R^n. also the intersection of A and its complement is empty. now if the R^n equal to union of A & B, then B must be complement of A. B is open but complement of A is closed. contradiction
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    Re: Complement of nonempty open set

    Quote Originally Posted by ansarey View Post
    the complement of open set is closed. so union of A and its complement equal to R^n. also the intersection of A and its complement is empty. now if the R^n equal to union of A & B, then B must be complement of A. B is open but complement of A is closed. contradiction
    There are typologies in which there are sets which are both open and closed.
    They are often called clopen sets.

    If fact, a space is connected if and only if the only clopen sets are the empty set and the whole set.

    In your proofs, you must use the properties of \mathbb{R}^n.
    You have done that so far.
    Thanks from ansarey
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    Re: Complement of nonempty open set

    dear friends, please introduce some ref about set, vector space, and especially properties of spaces (i.e connectivity, etc). i am phd student of engineering and working on dynamical system and nonlinear control. i am familiar with subject like open, closed, closure, clopen, etc. but i think i dont have enough insight and intuition in this field. for example, i can think about above problem only in level of definition (and try to prove it by searching a contradiction), and can not think in higher level (think about connectivity, and other properties of spaces like R^n as Mr plato and ModusPonens can do) to prove or understand this type of problems.
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    Re: Complement of nonempty open set

    Quote Originally Posted by ansarey View Post
    for example, i can think about above problem only in level of definition (and try to prove it by searching a contradiction), and can not think in higher level (think about connectivity, and other properties of spaces like R^n as Mr plato and ModusPonens can do) to prove or understand this type of problems.
    The book Elementry Theory of Metric Spaces by Robert B Reisel is written for a Moore style course. That makes it ideal for self-study. Do not let the title fool you. I used it as a text for a course in topology designed for pre-university teachers. The subtitle is "A Course in Constructing Mathematical Proofs.
    Last edited by Plato; November 24th 2012 at 03:59 PM.
    Thanks from ansarey
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