In this article we will explore the different aspects related to Talk:Perfect set, delving into its importance today and its relevance over time. From its origins to its impact on today's society, we will analyze the many facets of Talk:Perfect set and its influence in various areas, such as culture, economics, politics and daily life. Through a multidisciplinary approach, we will examine how Talk:Perfect set has evolved and adapted to the changes of the modern world, and how it continues to be a topic of interest and debate today. Through detailed and critical analysis, this article seeks to shed light on the many aspects of Talk:Perfect set and its implications for the present and future.
| This article is rated Start-class on Wikipedia's content assessment scale. It is of interest to the following WikiProjects: | |||||||||||
| |||||||||||
Is this correct? An isolated point forms itself an open set. How does that affect whether every point is a countable intersection of open sets?--Patrick 15:31, 13 June 2007 (UTC) I'm sure it's wrong. An open set is certainly a g-delta and so a discrete space has all points g-delta even though it is as far from perfect. I'll remove itA Geek Tragedy (talk) 16:27, 30 May 2008 (UTC)
Most of what I am going to say here is not in the true/false dichotomy, but is a matter of perspective, and therefore open to some degree of interpretation. The big picture is that I think the article misses the main point. The article lists many properties and non-properties about perfect spaces. But what is the main point about perfect spaces? As far as I understand (and I do not claim my understanding to be complete) the main point is that there exist totally disconnected perfect spaces. Like other results in point set topology, this has to do with exploring how bizzare things can get. (The standard example of a perfect space that is totally disconnected is not so very exotic, just our familiar Cantor set.)
In fact, the set of all rationals with the standard subspace topology (from the reals) and the set of all real numbers with the lower limit topology are both perfect and are totally disconnected. Also, there is nothing bizzare about the fact that there are totally disconnected spaces that are perfect (in my opinion). This is why I emphasized that perfect spaces aren't really 'perfect' so technically I did explore what you suggested.
Topology Expert (talk) 04:29, 6 July 2008 (UTC)
As far as I am aware, the notion of perfect space is not "useful" like the concept of compactness. By this I mean that I don't think it ever happens (or perhaps very seldomly) that you would prove that a space is perfect as a step in order to prove something else.
No. You wouldn't do this but sometimes proving a space is perfect can be used as a 'preliminary'. For example, before verifying that a space is simply connected, you would first verify that it is path connected before checking its fundamental group.
Topology Expert (talk) 04:29, 6 July 2008 (UTC)
The lede is very problematic. Usually, the lead should explain the main point of an article informally without going into technicalities and details. But in this case the lede is misleading. For me "generally large enough" is not any imprecise approximation to some fact about perfect spaces. The same applies to "more likely to be connected". Perhaps the sentence "Note however, perfect spaces are generally not substitutes for spaces satisfying other important topological properties" is meant to say something similar to what I have written in the last sentences of the previous paragraph, but that's really not clear to me.
A non-perfect space is more likely to be disconnected (therefore a perfect space is more likely to be connected). This is because a non-perfect Hausdorff space with more than one point is disconnected as indicated in property 6 (note that just adding the Hausdorff condition makes it disconnected along with the condition that the space must contain more than one point). The use of 'more likely' is quite vague but still gives the reader an impression that perfect spaces are useful in some respect (which is true). Also, the name 'perfect' can be misleading and can compel people to believe that perfect spaces satisfy 'many' useful topological properties. By writing what I did in the introduction, I am showing that this is in general not the case.
Topology Expert (talk) 04:29, 6 July 2008 (UTC)
In item 6, the assumption that X contains more than one point is missing.
I admit I made a mistake here. Therefore, I fixed this.
Topology Expert (talk) 04:29, 6 July 2008 (UTC)
Instead of Theorem 1, it is reasonable just to say "By the Baire category theorem a perfect compact Hausdorff space is uncountable."
Actually, theorem 1 doesn't follow from the Baire category theorem but the proof is similar to the Baire category theorem. Note that both theorems use the finite intersection property but this doesn't imply that they are dependent on each other.
Topology Expert (talk) 04:29, 6 July 2008 (UTC)
Theorem 2 is not really a theorem. It is an obvious fact.
Yes. It is an obvious fact and perhaps it would be better to include it in the 'examples and properties' section.
Topology Expert (talk) 04:29, 6 July 2008 (UTC)
Oded (talk) 05:45, 5 July 2008 (UTC)
Perhaps more important than all of my comments above: can you provide citations indicating that all these facts are known and meet the WP:V and WP:RS standards? Oded (talk) 18:01, 5 July 2008 (UTC)
Dear Oded,
I am supplying the proof of each and every property (the proofs in general lack minor details but should serve as verifications of each property)
Proof of 1:
The proof is already supplied in the article.
Proof of 2:
Since every subspace of X is perfect, so are two point sets (since X has more than one point, there are two point subsets of X). Therefore, since a two point set is not perfect, it must have the indiscrete topology so that one point sets cannot be closed.
Proof of 3:
Suppose a one point set is open relative to the product of arbitrarily many perfect spaces. Since the projection maps are open (with respect to both the box and product topologies), each co-ordinate of this point must be open relative to the respective space; a contradiction since each component space is perfect.
Proof of 4:
The proof is already supplied in the article.
Proof of 5:
The proof is already supplied in the article.
Proof of 6:
If X is a non-perfect Hausdorff space, then there is at least one isolated point in X. Since one-point sets are closed in X, this isolated point is also closed. Since there is a subset of X which is both open and closed (the isolated point), and since X has more than one point, X is disconnected. The proof is exactly the same if X is a non-perfect, T_1 space with more than one point.
Proof of 7:
An infinite set in the indiscrete topology is not Hausdorff and therefore cannot be metrizable; a space in the discrete topology is metrizable (the discrete metric induces the topology of this space) and is clearly not perfect.
Proof of 8:
A finite subspace of a T_1 space always inherits the discrete topology and is therefore not perfect.
Proof of 9:
If X is a topological space and if {x} is an isolated point of X, then X is locally connected at x. Similarly, X is locally compact at x. The rationals is perfect but not Baire; the set of all real numbers in the lower limit topology is perfect but not second countable; R^(w) given the box topology is perfect but not first countable.
Proof of 10:
The only metric topology on a finite point set is the discrete topology and the discrete topology is not perfect.
Proof of 11:
If G is a topological group having at least one isolated point (call it x), then consider the map f(z) = a*z for some element 'a' in G. It is a standard fact that 'f' is a homeomorphism (see Munkres, chapter 2 in the section entitled supplementary exercises on topological groups). If 'y' is an element of G then let a = y*(x^(-1)). Then f(x)=y and since f is a homeomorphism, {y} must be open.
Proof of 12:
If (G_a) is an open cover of (X, T) [where T is the indiscrete topology on X), then the collection (G_a) must contain X (the whole space) so that X itself is an open refinement of order 1. Therefore, the topological dimension of X is 0.
I have already given proofs of the theorems. I am quite sure that these proofs are correct but if you do come across any mistakes please let me know. Please also ignore the fact that in some proofs, some details may be left out.
Topology Expert (talk) 05:11, 6 July 2008 (UTC)
Dear Oded,
Please read what I have written in response to your comments. I have also given proof of every single fact. This should serve as a verification.
Thanks
Topology Expert (talk) 04:29, 6 July 2008 (UTC)
For every let . Since is perfect, is dense. Since is Hausdorff, is open. If would be countable, then the Baire category theorem would imply that is dense in . But , which is not dense in . (I guess it is taken for granted that is not allowed. Perhaps this should be made explicit.) QED Oded (talk) 17:01, 6 July 2008 (UTC)
Dear Oded,
Your proof is correct and I guess I didn't believe you earlier. Perhaps you should include this proof in the article. Before creating an account, I used to read Wikipedia for definitions in mathematics and quite often I did come across the use of iff so this is why I included it in the article. I also didn't write that perfect spaces are 'generally large enough'; I think Jack wrote that.
Thanks
Topology Expert (talk) 01:46, 7 July 2008 (UTC)
Also, note that if a space has 'more' (which cannot be made mathematically precise in the context of infinite sets unless you talk about countability and uncountability) open sets, it is more likely to be Hausdorff and less likely to compact. It is also less likely to be perfect. Basically this gives a rough idea to the reader of what types of spaces are perfect (the ones with fewer open sets although this doesn't guarentee that they are perfect). I guess that perhaps it is not necessary to include this in the article but this can be easily removed.
Topology Expert (talk) 01:50, 7 July 2008 (UTC)
I got to go now, but have a look. Never mind, though, it really does not matter who wrote this. More later. Oded (talk) 01:55, 7 July 2008 (UTC)
I am concerned that the definition of perfect space used in this article is not the correct (i.e., standard) one. Elsewhere on wikipedia are the following two terms:
All references for "perfect space" I have been able to find in the literature (e.g. Engelking, General Topology, p. 48) mean a space for which every closed subset is a G-delta set. (Thus perfectly normal = perfect + normal. Perhaps this was the origin of the term.) Since any metrizable space is perfectly normal, this is obviously an inequivalent use of the term. Moreover, I have certainly heard "Let X be a topological space without isolated points" many, many times: i.e., it is not deemed necessary to give this concept a special name.
Therefore I submit that the present article is devoted to a sequence of facts about spaces without isolated points, whereas it should rather be discussing the (somewhat less immediately transparent) concept of spaces in which every closed subset is a G-delta set.
If someone disagrees, please give references for this use of perfect space: since I can find many references for the other use of perfect space, more than one or two would be preferred. Plclark (talk) 16:04, 11 July 2008 (UTC)Plclark
Searching for perfect space on MathSciNet yields 97 hits. A cursory inspection seems to indicate that at least half of these refer to the "closed sets are G-delta sets" definition. I did not find any which clearly refer to the "no isolated points" definition although of course one would have to look harder to be sure. So as far as I can see the above assumption may not be warranted. I am not a general topologist, though; it might be nice to have one weigh in on the matter. Plclark (talk) 21:15, 11 July 2008 (UTC)Plclark
Dear all,
According to Rudin's 'Principles of mathematical analysis' (third edition), page 32, definition 2.18 (h), a subset E of a metric space X is perfect if E is closed and if every point of E is a limit point of E (i.e E has no isolated points). Note that in a metric space, every closed set is a Gδ set.
In particular this implies that the definition you found is not equivalent to Rudin's definition for metric spaces (Rudin requires the additional condition that the metric space has no isolated points; according to your definition every metric space is perfect). However, the definition in Rudin's does imply yours, i.e Rudin's definition is stronger that yours for metric spaces.
I just had a glimpse at chapter 2 of Rudin's book and I noticed that many of his proofs about compactness in metric spaces inexplicitly use the fact that metric spaces are Hausdorff (for instance theroem 2.34, page 37). In fact most of his theorems about compactness do generalize to arbitrary Hausdorff topological spaces. Therefore, I am led to believe that despite the fact that Rudin only considers metric spaces in the definition of perfectness, his definition probably does apply to arbitrary topological spaces.
Note also that I was not the one to give the definition of a perfect space; I expanded upon the definition.
The creator of the article defined a perfect space as a topological space that has no isolated points. I think that it is best to question the creator as to why he defined perfect spaces in this way. Most probably, he referred to Rudin's book when giving the definition. Topology Expert (talk) 11:11, 17 July 2008 (UTC)
The issue is most likely just that "perfect" is used to mean many things. Two books that use the same terminology as this article are Classical Descriptive Set Theory, Kechris, and Basic Set Theory, Levy. You can search for "perfect Polish space" on google books to find many more examples. Since a Polish space always has the property that a closed set is Gδ, the adjective perfect there always refers to the lack of isolated points.
I'll have to wait until tomorrow to look for references on spaces where every closed set is Gδ. I am pretty confident that Counterexamples in Topology has some relevant examples, it may have a term for them. — Carl (CBM · talk) 22:59, 3 August 2008 (UTC)
Let me return to the first point of this section:
Strangely, "perfect set" still redirects to "perfect space", and "perfect space" does not mention perfect sets at all. It that terminology obsolete? Boris Tsirelson (talk) 08:30, 18 August 2013 (UTC)
A quote from talk:WikiProject_Mathematics of 21 November 2013 (Section "Perfect set versus perfect sets"):
Perfect set redirects to Perfect space, while Perfect sets redirects to Derived set (mathematics). Rather strange. Boris Tsirelson (talk) 08:34, 20 November 2013 (UTC)
Boris Tsirelson (talk) 20:47, 21 November 2013 (UTC)
The result of the move request was: 'Closed as move. This page was moved during the move request, which should not have occurred per WP:NOMOVE. However, given that the listing period is now over and there are effectively three supports with no opposes, it seems fine to keep the new name. (non-admin closure) — Amakuru (talk) 14:17, 2 December 2013 (UTC)
Perfect space → perfect set – In my experience, the most usual definition one comes across for this notion is, S is perfect if S is closed and has no isolated points. Now, of course, every topological space is closed in itself, so if you insist on treating the notion as an intrinsic one about a whole space, instead of as a property of subsets of a space, then the "closed" part trivializes.
But that trivialization leads to pathologies. For example, the rational numbers with their usual topology are a "perfect space" in that sense (or a perfect subset of themselves, in what I argue is the more usual nomenclature). But that sort of perfect-ness has no connection with the reasons that perfect sets have historically been considered important.
The historical context for the notion is things like the Cantor–Bendixson theorem, which shows in particular that closed sets (of reals) cannot be counterexamples to the continuum hypothesis, because they can be decomposed uniquely into a countable set and a perfect set, and a nonempty perfect set must have the cardinality of the continuum. That's completely lost for pathological examples such as the rationals-in-themselves.
More modernly, perfect sets are important as the sets of branches of perfect trees, which are trees that always branch past any given node. Again, that has little connection with this intrinsic "perfect space" notion.
So my proposal is: Move the present article to perfect set, and reword the introduction to frame it as a property of subsets of a given topological space (closed and having no isolated points). Then mention the notion of perfect space, which is simply a space that is perfect as a subset of itself, and point out that this is the same as having no isolated points. Trovatore (talk) 19:44, 22 November 2013 (UTC)
I have rashly gone ahead and moved it. I haven't yet re-written the introduction, nor dealt with links to the new redirect page. Michael Hardy (talk) 22:23, 23 November 2013 (UTC)