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Math Help - Clarify def of closed set: implication?

  1. #1
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    Question Clarify def of closed set: implication?

    Is the definition of a closed set an implication?

    My understanding of the definition is this:

    OPEN: For a set U in R. U is open if for all x in U there exists an m>0 s.t. (x-m, x+m) is also in U.

    CLOSED: For a set F=R\U. If F is open then U is closed.

    So if I find that F is not open, does that mean that U is not closed?

    Thanks a bunch!
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  2. #2
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    Hello yvonnehr!

    Quote Originally Posted by yvonnehr View Post
    Is the definition of a closed set an implication?

    My understanding of the definition is this:

    OPEN: For a set U in R. U is open if for all x in U there exists an m>0 s.t. (x-m, x+m) is also in U.
    Yes, that's right


    Quote Originally Posted by yvonnehr View Post
    CLOSED: For a set F=R\U. If F is open then U is closed.

    So if I find that F is not open, does that mean that U is not closed?
    If F is open, then F^c is closed (^c is supposed to mean 'complement of')

    So if F is not open, I assume F is closed. What is F^c? It is U. So U is open.

    But: R and \emptyset is both open and closed.

    Does it answer your question?

    Kind regards
    Rapha
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  3. #3
    Junior Member nimon's Avatar
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    Alternative definition

    You might also find it helpful to know that you can characterize closed a set by knowledge of the limits of its sequences. Let (X,d) be a metric space, then F\subset X is closed if and only if, whenever the sequence (x_{n})_{n\in\mathbb{N}} \in X converges - to x, say, - then it follows that x\in F. I.e.: F is closed if and only if it contains all of its limits. You can prove this using your definition.
    Last edited by nimon; October 2nd 2009 at 02:17 AM.
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    Quote Originally Posted by Rapha View Post
    So if F is not open, I assume F is closed. What is F^c? It is U. So U is open.
    Be very careful with statements such as that.
    A door is either open or closed. That ain't true for sets.
    The set [0,1) is neither open nor closed.
    Its complement (-\infty,0)\cup [1,\infty) is also neither open nor closed

    If is true that a set is closed in and only if its complement is open.

    As has already been pointed out, the standard metric space definition of closed sets is:
    A set is closed if and only if it contains all of its limit points.
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  5. #5
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    Smile closed set-biconditional statement

    Quote Originally Posted by Plato View Post
    Be very careful with statements such as that.
    A door is either open or closed. That ain't true for sets.
    The set [0,1) is neither open nor closed.
    Its complement (-\infty,0)\cup [1,\infty) is also neither open nor closed

    If is true that a set is closed in and only if its complement is open.

    As has already been pointed out, the standard metric space definition of closed sets is:
    A set is closed if and only if it contains all of its limit points.
    This makes sense to me. So, the definition is a biconditional statement rather than an implication. I just wanted to make sure so that I can use is properly in a proof. Thank you so much, Plato.
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