Talk:Polish space

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Template:Maths rating

Given that the topology glossary does not in fact include a definition of a polish space, it seemed rather silly to replace this article with a redirect. —The preceding unsigned comment was added by (talkcontribs) 10:43, 26 February 2005 (UTC)

Each Polish Space is a Baire Space

according to the Wikipedia article on Baire Spaces. So "Baire Space" cannot be an example for a Polish Space.

I'm unfamiliar with both notions, so I won't edit the page myself... —The preceding unsigned comment was added by (talkcontribs) 10:03, 9 November 2006 (UTC)

You've run into an unfortunate ambiguity. Some mathematicians use the phrase "a Baire space" to describe any space satisfying the Baire category theorem. Others use the term "Baire space" (no "a") to mean one particular space (namely all functions from the naturals to the naturals, given the product topology). In this article, the second sense is intended. --Trovatore 19:46, 9 November 2006 (UTC)

Typesetting problem

I've just corrected G\delta to Gδ. I've tried at first

[[G-delta_set|<math>G_\delta</math>]] subset

which produces subset. Why this is working wrong? --Kompik 15:49, 18 January 2007 (UTC)

Well, it seems alright now. I should have made a screenshot -- otherwise nobody believes me that when I've included it for the first time, it has produced something strange. :-) --Kompik 06:30, 22 September 2007 (UTC)

Converse of Alexandrov's Theorem

I've stumbled upon this result in Arveson. It was mentioned without proof and Bourbaki was included as the reference, therefore I've included both these books into references. —The preceding unsigned comment was added by Kompik (talkcontribs) 18:19, 28 January 2007 (UTC).

This is also proved in Kechris thm 3.11 Miaoku (talk) 15:56, 19 November 2007 (UTC)

Confusing statement in the article

This statement doesn't make sense to me:
(Cantor-Bendixson theorem) If X is Polish then any closed subset of X can be written as the disjoint union of a perfect subset and a countable open subset.
By this statement, the set Q of the rationals (i.e. a closed subset of the reals) is the disjoint union of a perfect subset and a countable open set. But there are no countable open subsets in the reals, so this implies that Q is a perfect subset (i.e. a closed set with no isolated points). This is false because every point in Q is an isolated point. Timhoooey 03:02, 14 October 2007 (UTC)

Good catch; the "open" was nonsense (I guess it's open in the relative topology induced on the subset, but that's not how the statement would ordinarily be read). However your example is wrong -- Q is not closed, and no point in Q is isolated. --Trovatore 03:07, 14 October 2007 (UTC)
Oops... you're right. 5+ straight hours of studying measure theory will do that to you. I was thinking that countable implies all isolated (as in the naturals) and that Q is closed in itself, not in the reals. Time to take a break. Timhoooey 03:50, 14 October 2007 (UTC)

Definition of a Lusin space

Out of my element here, but should it not be "X is Lusin if it admits a finer Polish topology" instead of a "weaker" Polish topology? —Preceding unsigned comment added by (talk) 13:08, 23 August 2009 (UTC)