Talk:Baire category theorem

message: Can someone reference the connection to the axiom of choice? Gady 21:58, 12 Nov 2004 (UTC)
thread_title:  Axiom of choice 

message: This article needs a naive statement of the theorem too. It should be possible for someone to look at this article and get something out of it without having to reference the Baire Space article. Ewjw 17:44, 8 August 2006 (UTC)
replies: text: Yes, IMHO the article is needlessly confusing for readers who are coming here just for the Baire Category Theorem for complete metric spaces (which are presumably most readers). I propose removing most of the references to Baire spaces and mentioning it as a generalization. Functor salad 18:40, 27 September 2007 (UTC)
thread_title:  Too technical! 

message: In Section "Statement of the theorem", is the example given in the last paragraph correct?
:I think an explanation is hinted at in the first paragraph of the section The set of all irrationals in irrational number. It depends on the metric; the space of irrational numbers becomes complete if you use a nonstandard metric. I'm not sure of this though; hopefully somebody can confirm this. -- Jitse Niesen (talk) 04:08, 9 October 2006 (UTC)
thread_title:  Irrational numbers complete?? 

message: may i suggest that section undergo some expansion. although it's obvious that the axiom of countable choice is required in the usual proof of Baire category theorem, would be good to point that out. a few remarks about why Baire category theorem implies axiom of dependable choice, if doable, might be good also. Mct mht 08:09, 31 May 2007 (UTC)
thread_title:  axiom of choice 

message: Doesn't BCT3 need to read "A non-empty complete metric space is NOT the countable union of nowhere-dense CLOSED sets"? At the moment it doesn't say the sets need to be closed.
:I don't think you need to assume the sets are closed in addition to being nowhere-dense (or, equivalently, has dense complement.) See  for instance; the page gives a number of equivalent formulations. -- Taku (talk) 23:09, 4 January 2009 (UTC)
replies: text: The problem comes with the definition of nowhere dense. In the Nowhere dense set article it is said that the interior of the closure is empty.  In that case you obviously don't need to have closed sets, since you may close them anyhow.  But in the present article it is said the complement is dense, and then  BCT3 becomes obviously wrong, as the example of  rationals and irrationals show: they are both nowhere dense in the second sense (which is not correct, in my opinion), but the union is the complete metric space R.--Bdmy (talk) 11:11, 5 January 2009 (UTC)
replies: text: Another issue: the definition of Baire space is that the intersection of a sequence of dense open sets is dense.  It is not equivalent to say that the intersection is not empty, and BCT3 as stated today is just that second statement.  Therefore this form of BCT3 is not equivalent to BCT1. --Bdmy (talk) 11:18, 5 January 2009 (UTC)

text: Example: take X to be the union of [0, 1] and all the rational points in (1, 2), with the topology induced by the real line.  Then the intersection of a sequence of dense open subsets of X is never empty, but might fail to be dense in X.  --Bdmy (talk) 11:36, 5 January 2009 (UTC)
replies: text: I meant: BCT3 is not equivalent to being a Baire space.  If you know the space is complete metrizable, then both properties are just true, and the word equivalent is meaningless. --Bdmy (talk) 12:06, 5 January 2009 (UTC)
thread_title:  BCT3 wrong? 

message: The proof section of the article appears to have numerous flaws.  For example, it starts as follows: 
:Let U_n be a collection of open dense subsets. We want to show that the intersection \bigcap U_n is dense. For that, let W \subset X be an open subset. By denseness, there is x_1 and r_1 > 0 such that:
::\overline{B}(x_1, r_1) \subset W \cap U_1.
The above fails to define what \overline{B}(x_1, r_1) is -- it could be anything, but a clever reader might guess that its a ball centered on x_1 or radius r_1 -- but one shouldn't have to be a clever guesser. What about the overline? is this an open or closed ball?  Other proofs seem to require only open balls, why is a closed ball used here?  Is this an important difference?
Much worse is the problem: "open ball in what space?" Consider, for example, the topological space X == the real numbers, and the sets U_n == the rationals. i.e. U_n=U_m for all m,n, so each of the U_n is the same set (i.e. is the rationals). As far as I know, the Baire category theorem applies to this case, right? But here we have a problem:  There is no open ball in the reals that is a subset of the rationals, and so the proof breaks down.  So maybe the open ball was supposed to be a ball in U_1, and nevermind X? 
:Ahh, silly me, the rationals are not open in the reals, so the premise of the proof is violated. But still, I found the proof to be confusing -- it should state: "an open ball in X that is a subset of U_1 cap W" just to crisp things up. Otherwise, the naive reader is left guessing as to the details. 
The lack of clarity here makes this much more impenetrable than it needs to be. linas (talk) 16:33, 2 July 2009 (UTC)
replies: text: I agree that it's better to explain what B and \overline{B} mean here, which the article does now. A closed ball instead of an open ball is used here because that would facility the argument at the end: the limit is in both W and all U_n. -- Taku (talk) 22:43, 2 July 2009 (UTC)
thread_title:  Flawed proof 

message: The derived set article says "Perfect sets are particularly important in applications of the Baire category theorem" and links here. However, there is no mention of perfect sets in this article. As near as I can tell, the quoted sentence just means that perfect sets are often complete metric spaces themselves and so BCT1 applies to them. (Perhaps BCT2 often works as well? My topology is not very deep.) For instance, perfect subsets of R are closed and any Cauchy sequence can be bounded, so completeness follows from compactness, and so all perfect subsets of Euclidean space are Baire. However, Q is perfect (in itself), but not Baire. In any case, some discussion of perfect sets would be nice.
On another note, I strongly dislike the organization of this article. It discusses three related but different theorems simultaneously and so feels schizophrenic. Perhaps it could be improved by having three separate sections on each version of the theorem. That would also put the most common one, the complete space version, first and foremost where it belongs. Also, you have to flip back and forth between this page and the Baire space article. If the two were merged, this article's theorems could just become their own section of that article. There's already a sizable summary of this page on that one, and the examples could probably be made to flow well if the two were merged. 24.220.188.43 (talk) 19:43, 8 May 2011 (UTC)
thread_title:  Perfect sets 

message: It would be nice to add an informal / visual explanation why the axiom of countable choice does not suffice (beside just saying "BCT 1 is equivalent over ZF to the Axiom of dependent choice"). Mathelerner (talk) 21:32, 25 July 2022 (UTC)
thread_title:  Axiom of Countable choice