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An set without limit points - Necessarily closed?

  1. Aug 13, 2007 #1
    Definitions:
    "x is a limit point of A" = "All neighborhoods of x contain an infinite amount of points of A"
    "x is a contact point of A" = "All neighborhoods of x contain at least one point of A"
    "X is a centered system of closed sets" = "[tex]\cap A[/tex] is not empty, where A is any finite subset of X, and where X is a set of closed sets."

    In some book of mine, in some proof of a theorem, the author implicitly asserted for some set X that "Since X is a set without limit points, X is closed."

    Now, I do not really know how to prove that--in fact, I think it may be false.

    For example take the space to be the set of integers. Let the open sets be any set of non-negative integers, sets of the form {a, -a} where a is any natural number, any unions of the above sets, and the empty set. Let N be the set of natural numbers. Surely N is without limit points--in fact, no element of the space is a limit point of any set since all elements of the space have a finite neighborhood. However, surely -1, which is not a an element of N, is a contact point.

    Note: this is what the author stated:

    "Theorem: If T is a compact space, then any infinite subset of T has at least one limit point.
    Proof: Suppose T contains an infinite set X with no limit point. Then T contains a countable set [tex]X=\{x_1,x_2,...,x_n,...\}[/tex] with no limit point. But then the sets [tex]X_n=\{x_n,x_{n}_{+1} ,...\} (n=1, 2, ...)[/tex] form a centered system of closed sets in T with an empty intersection, i. e. T is not compact."
     
    Last edited: Aug 14, 2007
  2. jcsd
  3. Aug 14, 2007 #2
    A set is closed if its complement is open, now if X every point of it is not a limit point, then obviously the limit points of a sequence of points in X, are in X^c, now you should show, that X^c contains an open sphere, which is basically proving that there exists a point P_n such that for every r>0, and every P in the discussion set, we have that: |P-P_n|<r, which is the definition of a limit point.
     
  4. Aug 14, 2007 #3

    morphism

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    If a set is closed it contains all its limit points, or stated differently, a set isn't closed if it doesn't contain at least one of its limit points. So if a set has no limit points, it must be closed.
     
  5. Aug 14, 2007 #4
    "If a set is closed it contains all its limit points" but the converse true?
    For example, consider the two-element space T= ({0, 1}, {{}{0}{0,1}}). {0} does not have any limit points, but it is not closed, since {0,1}-{0}={1} is not an open set.
     
    Last edited: Aug 14, 2007
  6. Aug 14, 2007 #5

    matt grime

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    This is clearly wrong: The integers as a subset of R with the usual topology is not closed and has no limit points.
     
  7. Aug 14, 2007 #6
    Actually, the integers do form a closed set in R with the usual topology, since it contains all its contact points (or, formulated another way, the union of the sets (a, a+1) where a is any integer is an open set, which is the complement of the integers.)
     
  8. Aug 14, 2007 #7

    matt grime

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    This is false.

    I assume that the X above is related to the X below.

    The sets X_n are not a priori closed: countable unions of closed sets are not guaranteed to be closed. I can only presume that the author has some other result in mind that utilizes the fact that T is compact.
     
  9. Aug 14, 2007 #8

    WWGD

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    But, if a set S did not have any limit points,and it was not closed, you may
    say that the set of limit points of S is the empty set. Then saying that S is
    not closed would imply that S does not contain the empty set, which cannot happen.

    In addition, as I think someone above said, if S did not have limit points,
    then you may consider X-S (X ambient space, assuming S is a subspace).
    Then, for any x in X-S , there must be a 'hood ('hood=neighborhood) V_x
    of x, such that V_x does not intersect S , otherwise x would be a limit
    point of S . That gives you, for every x in X-S , a 'hood of x contained
    in X-S . Then X-S is open, and S is closed.
     
  10. Aug 14, 2007 #9

    WWGD

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    Actually, 1 is a limit point of {0}, since every 'hood (neighborhood) of
    1, specifically {0,1}, intersects {0}.
     
  11. Aug 14, 2007 #10
    I believe that we should stick with proper english, and not with english from the hood.
     
  12. Aug 14, 2007 #11
    But with the definition of limit point being "a point which contains an infinite amount of points in A in all its neighborhoods," no set in that space can have a limit point, since they are all finite sets. Perhaps the statement would work if the definition was replaced by "a point x which contains at least one point in A-{x} in all its neighborhoods" but it does not work with the current definition of limit point.
     
  13. Aug 14, 2007 #12

    WWGD

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    To extropy:

    Well, I am going by Borowski and Borwein's Dictionary of math's def:

    Cluster point, limit point, accumulation point:
    "A point every punctured neighborhood of which has a non-empty intersection with a given set; ......". and no xception is made for finite spaces. I guess it comes down to the definition you're using. So,
    to sum it up, I am right under my definition, and you are right under
    yours, I believe.
    But I don't know which one we should use. Problem is if we do not have limit points, we cannot talk about convergence, etc. , so that neither of the sequences : 0,0,0,..... nor 1,1,1,..... would converge.



    To Loop Quantum Gravity:
    I was just joking a little, just for fun. I am not writing entire messages
    in that style, I just like to drop jokes like that everyonce in a while. Math
    can get too intense at times, and I thought that may help.

    If I was writing fully in that style, or if most of my posts were not
    relevant to the questions, I would agree with you. Still, I guess this
    is a democracy , and if many people dislike it, I will drop it. Sound fair?
     
  14. Aug 15, 2007 #13
    I don't have a problem with it, but if it were for short or fun then why have you used it with brackets that it means neighbourhood more than once? (-:
     
  15. Aug 15, 2007 #14

    matt grime

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    Just an FYI, the normal abbreviation is nbd.
     
  16. Aug 16, 2007 #15

    WWGD

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    To Quantum Loop:
    Well, I decided to define it twice because I did not know wether everyone would read both my posts, so someone who had no read the post where I first used the abbreviation would not understand it.

    To Matt:
    Yes, Nbd is the usual abbreviation, but using it would defeat (at least my) purpose of giving (at least myself) a relaxation break , by using something out of the ordinary.
    I don't want to go too far off, but usage of standard words triggers
    or reinforces existing moods. Using something different or off-beat may , at best, shock you out of a state of stress, and , at worse (or so I believe) it may be just a small distraction.

    Again, if this upsets people, I will drop it. Anyway, peace, and go with
    the force, go with the m*a (f=m*a) :)

    P.S: I wrote a title for a HW I did on Latin Squares: " I am Latin, but
    I ain't no square" ( I am Latin, from Venezuela, BTW). This is the
    ( I admit low-brow, low-quality) humor I am aiming for.
     
  17. Aug 17, 2007 #16
    it's dp/dt, but i let you off the hook. (-:
     
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