Talk:Derived set (mathematics)

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Why does perfect set redirect here?

A perfect set is not a synonym for a derived set, nor is perfect set closely tied to the terminology derived set. A perfect set is a closed set with no isolated points. The fact that you can express this in terms of derived sets is irrelevant.

Google "perfect set" you get the definition for derived set because of this idiotic redirect. Whoever is doing such things, please stop making Wikipedia unusable by deleting useful pages. I'm recreating the perfect set page. — Preceding unsigned comment added by (talk) 18:00, 29 March 2013 (UTC)

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). -- (talk) 15:32, 19 December 2007 (UTC)

We say that a point x in X is a limit point of S if every open set containing x also contains a point of S other than x itself (Wikipedia). Hence, in the example above, x is not a limit point of S={x}: there is an open set {x,y} containing x which does not contain a point from S={x} other than x itself. —Preceding unsigned comment added by (talk) 15:07, 5 August 2008 (UTC)

  • Hmm. That example also violates condition 3' (but not condition 3, which is claimed to be equivalent to it, but it looks like isn't). In fact conditions 1 and 3' make the derived set of any finite set empty. On the other hand, I think 1,2,3',4' are correct for T_1 spaces. --Unzerlegbarkeit (talk) 17:10, 13 August 2008 (UTC)

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. 17:41, 8 January 2007 (UTC)


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. 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)
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