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

  1. #16
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    Maybe.
    Hope I can find a counterexample that A' is not closed in the future. Thank you Plato, for your patient replies.
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  2. #17
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    A subset of a T0 space whose derived set is not closed

    Consider \mathbb{R} with the topology generated by \{(-\infty,a) \mid a \in \mathbb{R} \}, which is a T0 space. The derived set of \{0\} is (0, \infty), which is open.
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  3. #18
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    A beautiful counterexample! Thank you for your brilliant first-post.
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  4. #19
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    Quote Originally Posted by austinmohr View Post
    Consider \mathbb{R} with the topology generated by \{(-\infty,a) \mid a \in \mathbb{R} \}, which is a T0 space. The derived set of \{0\} is (0, \infty), which is open.
    Ack - a typo. The set (0, \infty) is neither open nor closed in this topology. The counterexample still works, however.
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  5. #20
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    Quote Originally Posted by austinmohr View Post
    Consider \mathbb{R} with the topology generated by \{(-\infty,a) \mid a \in \mathbb{R} \}, which is a T0 space. The derived set of \{0\} is (0, \infty), which is open.
    Quote Originally Posted by austinmohr View Post
    Ack - a typo. The set (0, \infty) is neither open nor closed in this topology. The counterexample still works, however.
    If this is Lynn Steenís Left Order Topology, then isnít it true that 0 is a limit point of \{0\}?
    If so, then is 0 in the derived set of \{0\}?
    If that is so, what is its complement?
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  6. #21
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    Quote Originally Posted by Plato View Post
    If this is Lynn Steenís Left Order Topology, then isnít it true that 0 is a limit point of \{0\}?
    If so, then is 0 in the derived set of \{0\}?
    If that is so, what is its complement?
    We may be using slightly different definitions. I use "derived set" to mean the set of all accumulation points of a set A. Hence, if A is singleton, it cannot be contained in its own derived set by definition.
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  7. #22
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    Quote Originally Posted by austinmohr View Post
    We may be using slightly different definitions. I use "derived set" to mean the set of all accumulation points of a set A. Hence, if A is singleton, it cannot be contained in its own derived set by definition.
    You are correct. I was not thinking of a singleton set.
    In fact, I don't know what I had in mind.
    Maybe ( - \infty ,0] I really don't know.
    Last edited by Plato; November 24th 2009 at 02:25 AM.
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