Talk:Derived set (mathematics)
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[edit] S** subset S*
This is not a property of the derived set operation: Consider the set {x,y} with the topology where only the whole set and the empty set are open. Take S={x}. Then S'={y} and (S')'={x}
I don't know how old this unsigned remark is, but: If S={x}, then S*={x,y} and not only {y}. ! Therefore we have that S** is a subset of S*, since S** and S* are actually equal (in this example). --131.234.106.197 (talk) 15:32, 19 December 2007 (UTC)
[edit] Relation to closure
What's the relation between the derived set of a set, and the closure of that set? It would seem that in metric spaces the concepts coincide, no? -GTBacchus(talk) 07:28, 4 December 2006 (UTC)
- A derived set might not contain some of the points of the original set, eg. isolated points. But the closure of a set is the union of the set with its derived set. 192.75.48.150 17:41, 8 January 2007 (UTC)
[edit] Bendixson
- Bendixson proved that every uncountable closed set can be partitioned into a perfect set, called the Bendixson derivative of the original set and a countable set.
The name "Cantor-Bendixson theorem" rings a bell, and I think it was something like this, but this isn't quite right is it? Take an uncountable set which is totally discrete. It is an uncountable closed set, as required. But the only perfect set is empty, and the remainder is uncountable. Maybe there's some condition missing here. Google doesn't help and I don't have access to my dead trees at the moment. 192.75.48.150 17:32, 8 January 2007 (UTC)
- Thanks for pointing that out. I fixed the statement to make it correct. I don't know if it really belongs here, however. CMummert 17:43, 8 January 2007 (UTC)
