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Math Help - Is derived set closed in any topological space?

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    Is derived set closed in any topological space?

    Given a metric space X and a subset A of X, it can be easily proved that derived set A'(set of all limit points of A) is closed. Does this hold if X is a general topological space? Thanks!
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    Quote Originally Posted by zzzhhh View Post
    Given a metric space X and a subset A of X, it can be easily proved that derived set A'(set of all limit points of A) is closed. Does this hold if X is a general topological space? Thanks!
    If x is limit point of a set A and \mathcal{O} is an open set such that x\in \mathcal{O} then  \left( {\exists y \in \mathcal{O}} \right)\left[ {y \in A \wedge y \ne x} \right]<br />
.
    Think about what that implies about a limit point of A'.
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    Quote Originally Posted by zzzhhh View Post
    Given a metric space X and a subset A of X, it can be easily proved that derived set A'(set of all limit points of A) is closed. Does this hold if X is a general topological space? Thanks!
    How did you define a closed set in a topological set? Did you say it was the complement of an open set; or the closure of the set is the set itself?
    Anyway, you can see that the idea of a limit point carries over into a topological space right?

    Oops, Plato beat me to it, and his explanation is better than mine...
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    Quote Originally Posted by Plato View Post
    If x is limit point of a set A and \mathcal{O} is an open set such that x\in \mathcal{O} then  \left( {\exists y \in \mathcal{O}} \right)\left[ {y \in A \wedge y \ne x} \right]<br />
.
    Think about what that implies about a limit point of A'.
    Let x be a limit point of A', U an arbitrary neighborhood of x. To prove that x\in A' so that A' is closed, we must find an element y in U satisfying y\in A and y\ne x. Since x is a limit point of A', there is in U a point x'\in A' and x'\ne x. The statement can be proved if the space is metric, or Hausdorff, since we can find a smaller neighborhood around x' that is contained in U and disjoint from x; the element y is in this smaller neighborhood due to the fact that x' is a limit point of A. But in a general space, I'm not clear if this smaller neighborhood can be found. Could you please give me a further detailed hint?
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    Quote Originally Posted by zzzhhh View Post
    Let x be a limit point of A', U an arbitrary neighborhood of x. To prove that x\in A' so that A' is closed, we must find an element y in U satisfying y\in A and y\ne x. Since x is a limit point of A', there is in U a point x'\in A' and x'\ne x.
    You are missing a key concept here.
    What you have above is: any limit point of A' is also a limit point A.
    In other words, \left( {A'} \right)^\prime \subset A' so that A' is a closed set.

    In a general space, the definition of limit point does change.
    Last edited by Plato; October 2nd 2009 at 01:04 PM.
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    Thank you Plato, but I cannot understand your post completely, do you mean: A is fixed, but in different topological spaces, its derived set A' is different, some are closed, the others are not, right? My question is: Is it possible that A' is always closed in all topological spaces? I can't prove it, nor can I find any counterexamples.
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    No A can be any set in any top-space.
    The anwer is yes. A' is closed.
    Any limit point of A' is also a limit point A.
    In other words, \left( {A'} \right)^\prime \subset A' so that A' is a closed set.

    See
    Last edited by Plato; October 2nd 2009 at 01:21 PM.
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    I found a clue in Sierpinski's "Introduction to general topology". It is asserted as a theorem that A''\subseteq A' (See Fig 1 below). But the proof denpends on the hypothysis that the space be Hausdorff (actually T_1 axiom, the lowest level of separation axiom, on which the infinity property can be satisfied, see Fig 2 below). But my question is: Is A' closed in all topological spaces (here "all topological spaces" means, of course, all topological spaces containing A, sorry for omitting this qualifier in my last post), which do not necessarily satisfy the T_1 axiom?



    Fig 1. extracted from Sierpinski's "Introduction to general topology", Page 29-30.



    Fig 2. extracted from Munkres' "Topology" Page 99. (both books are available online from e.g. wikipedia)
    Last edited by zzzhhh; October 3rd 2009 at 06:23 AM.
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    Why do you think that any open set containing a limit point of a set must contain infinitely many points of that set? There is nothing in the definition of limit point that even implies that.
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    Quote Originally Posted by Plato View Post
    Why do you think that any open set containing a limit point of a set must contain infinitely many points of that set? There is nothing in the definition of limit point that even implies that.
    I didn't think that, this holds only when T_1 axiom are satisfied, that's why I couldn't prove it. which post of mine gives u that impression? I repeatedly stress the question is in a general topological space.

    The anwer is yes. A' is closed.
    Any limit point of is also a limit point .
    In other words, so that is a closed set.
    OK, show me the proof. Note that T_1 axiom is not assumed.
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    Suppose that k is a limit point of the derived set T’, i.e. k\in (T')'.
    Suppose that k is not in \overline{T} the closure of T.
    For any open set \mathcal{O} if k\in \mathcal{O} then \left( {\exists j \in T'\cap \mathcal{O}} \right)\left[ {j \ne k} \right].
    But that means j is a limit point of T.
    By the definition of limit point, j \in \mathcal{O}\, \Rightarrow \,\left( {\exists h \in T \cap \mathcal{O} } \right)\left[ {j \ne h} \right].

    So we have shown that any open set that contains k also contains a point of T that is different from k.
    By definition of limit point that shows that k is a limit point of T.
    In other words we have shown \left( {T'} \right)^\prime   \subseteq T'.
    T’ contains all of its limit points. It is closed.
    Last edited by Plato; October 3rd 2009 at 07:17 AM. Reason: Added one condition.
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    how do u exclude the possibility that h=k?
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    Quote Originally Posted by zzzhhh View Post
    how do u exclude the possibility that h=k?
    Add one line: suppose that k is not in \overline{T} the closure of T.
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    what if k\in T, or more precisely, \overline{T}<br />
-T'?
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    Quote Originally Posted by zzzhhh View Post
    what if k\in T, or more precisely, \overline{T}-T'?
    Yes, I now think that you do need at least T_1 space.
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