Talk:Alexandroff extension

Why restrict scope?

In a previous version of the article, the notion applied to any topological space. Now it is restricted to locally compact non-compact and Hausdorff space. I don't complain about requiring the space to be non-compact, but what is the reason for the other restrictions? Oded (talk) 23:55, 10 July 2008 (UTC)[reply]

I responded to this by giving the construction in the general case, which (following Willard's book for terminology) I called the "Alexandroff extension" of X. Possibly the name of the page should be changed to reflect that, and one-point and Alexandroff compactification should redirect here. Plclark (talk) 05:12, 11 July 2008 (UTC)Plclark[reply]

I removed the following assertion from the article:

"If the one-point compactification of X is homeomorphic to the one-point compactification of Y, then X is homeomorphic to Y"

This was added by User:Topology Expert. What is your source for this? How do you know it's correct? (In fact it is not true.) Plclark (talk) 06:19, 11 July 2008 (UTC)Plclark[reply]

I made many substantial revisions. Some further restructuring is necessary -- e.g., I am not yet sure what the name of the article should be! -- but I will let it sit for a while.

Also, I recently noticed that the term "extension" is used for a dense embedding of a topological space into another space. In order to be compatible with this usage, probably the construction should be limited to noncompact spaces and one should define the Alexandroff extension of a compact space to be itself. More later... Plclark (talk) 07:19, 12 July 2008 (UTC)Plclark[reply]

Rather trivial comment

Recently, User:EmilJ simplified a key formula in this article, namely replacing

$c^{-1}(c(X)\setminus (Y\cap c(X))$

with

$c^{-1}(c(X)\setminus Y)$

The only question here is: does $X\setminus Y$ imply (on wikipedia) that Y is a subset of X, or is it defined for all sets as $X\cap Y^{c}$ ? Plclark (talk) 22:56, 31 July 2008 (UTC)[reply]

I have never seen the notation

X\setminus Y

implicitly imply

Y\subset X

. Oded (talk) 23:32, 31 July 2008 (UTC)[reply]

I thought I had somewhere and was trying to be careful. Upon closer inspection, relative complements are discussed on wikipedia, at complement (set theory). Plclark (talk) 02:45, 1 August 2008 (UTC)[reply]

I think it is established that it is not enough to attempt consistent notation throughout WP, but whenever terminology or notation is not consistent in the literature, it is necessary to clarify. The point is that each article should not depend on other articles to clarify terminology, since WP is somewhat fluid. But in this case, I think there is no problem. Oded (talk) 03:14, 1 August 2008 (UTC)[reply]

Hausdorff?

In the section titled "The Alexandroff extension" I think we need to require X to be a Hausdorff space for the defined topology to actually be a topology. Consider the topological space X consisting of the real numbers with an additional point called 0'. Let the topology be generated by the base of open intervals of the reals, and open intervals of the reals in which 0 has been replaced by 0'. Then the interval [-1,1] is compact (but not closed) and the interval (-1,1) with 0 replaced by 0' is open, so {0'} is the intersection of two "open" sets in, X*, but it itself isn't "open". I may have made an oversight (or maybe to be as general as possible we should take these sets as a subbase?), but I'm convinced enough to add the Hausdorff requirement to the text for the time being (we can still safely drop local compactness as long as we don't care if the resulting topology is Hausdorff) Getspaper (talk) 17:12, 1 March 2009 (UTC)[reply]

This is page 140 of Willard, 19A. It gives the name "Alexandroff extension" precisely for the more general situation. I believe the result is also in Sierpinski's General topology. JackSchmidt (talk) 17:29, 1 March 2009 (UTC)[reply]

Ok, I undid my other edit adding the Hausdorff requirement, but are you sure the sources drop the requirement? If my "counterexample" above is wrong, I'd appreciate it if you could point out my error so I understand this better, thanks Getspaper (talk) 17:54, 1 March 2009 (UTC)[reply]

I am sure (Willard is reasonably explicit, "any space" and "X* is Hausdorff iff X is lcoally compact and Hausdorff"). Kelley's General Topology page 150 also has the result. I didn't find the Sierpinski page. I don't typically do general topology and the person with whom I was working on this article died recently, so it is a little depressing to go through it too carefully. Some of the surviving collaborators are probably better able to address it. User:Algebraist I think is quite comfortable with this stuff and he reads WP:RD/MATH (and probably this talk page). Oh and User:Plclark is also a good bet; I'm not sure if he reads the ref desk but almost certainly this talk page. JackSchmidt (talk) 18:15, 1 March 2009 (UTC)[reply]

I think that Getspaper is right. It is easy to see that the definition in the article is a topology if and only if every compact subset of X is closed (consider the intersection of the open sets X and

(X\setminus C)\cup \{\infty \}

for a compact C), and that's not true in general for non-Hausdorff spaces. If the Alexandroff extension is to be introduced for non-Hausdorff spaces, the topology must be defined in a different way. — Emil J. 18:54, 1 March 2009 (UTC)[reply]

Kelley additionally requires the complements to be closed (not just compact). This may be an error in Willard, or it may be some subtle point wrapped up in neighborhoods versus open sets. Both Kelley and Willard allow the starting space to already be compact. Both show that the resulting extension is compact and contains the original as a subspace. Does this reconcile the article, the sources, and the counterexample? JackSchmidt (talk) 19:10, 1 March 2009 (UTC)[reply]

I don't have any sources handy, but as far as I can see the definition in the article now works, and the counterexample no longer applies. I've fixed some omissions in the properties. — Emil J. 19:19, 1 March 2009 (UTC)[reply]

Yeah, I'm pretty sure requiring the complements to be closed as well as compact is enough to ensure both that X* is a topology and that it is Hausdorff if and only if X is locally compact and Hausdorff. Requiring X to not already be compact is only necessary for the embedding to be dense; and the map is always an embedding. But if a compactification is defined as a dense embedding into a compact space (I don't think I've ever seen another definition), then the lede should be changed so that it doesn't say the space has to be Hausdorff or locally compact for the extension to be a compactification, I think. Is that right? Getspaper (talk) 19:42, 1 March 2009 (UTC)[reply]

Oh, this is probably an old argument about whether compact means Hausdorff and quasi-compact or not. Kelley and Willard seem to like non-Hausdorff compact spaces, so with some small mention of the other convention, I think it'd be fine to expand the coverage. Did my recent edit address this? I think it is good to mention the "downside" to the A.e. but the old lead seemed to contradict the compactification article's (and getspaper's) reasonable definition. JackSchmidt (talk) 20:00, 1 March 2009 (UTC)[reply]

I was the one who wrote the article in its new form. Yes, the Alexandroff extension can be (and has been in many standard sources) defined on an arbitrary space, and yes, the open neighborhoods of infinity should be complements of closed, compact subsets of the original space. So the current version is correct. It may be that "closed" was missing in Willard's definition (I haven't gone back to check), but nevertheless the oversight is still mine. Thanks to all of you for catching it. Plclark (talk) 22:02, 1 March 2009 (UTC)[reply]

The one-point compactification (Hausdorff requirement?)

In the article Compactification (mathematics) a compactification of a topological space $X$ is defined as an embedding $i\colon X\to X^{*}$ where $X^{*}$ is (of course) a compact space and the image of $X$ is dense in $X^{*}$ . In the case of the Alexandroff extension $c\colon X\to X^{*}$ the image $c(X)$ is dense in $X^{*}$ if (and only if?) $X$ is non-compact, and since $c$ is an embedding this is sufficient (and necessary?) for $c$ to be a compactification. However in the section One-point_compactification#The_one-point_compactification is stated that it is a compactification if and only if $X$ is Hausdorff. I think this is incorrect, since it's not necessary (or sufficient) for $X$ to be Hausdorff. User:82.57.177.147 (talk), 14:54, 9 July 2014‎

Whaaaat?

The article currently states:

Put $X^{*}=X\cup \{\infty \}$ , and topologize $X^{*}$ by taking as open sets all the open subsets U of X together with all subsets V which contain $\infty$ and such that $X\setminus V$ is closed and compact.

On first reading, this sounds wrong, but after re-reading three times, I see that its right. Perhaps the following is more clear?

Put $X^{*}=X\cup \{\infty \}$ , and topologize $X^{*}$ by taking as open sets all the open subsets U of X together with all sets $V=(X\setminus C)\cup \{\infty \}$ where C is closed and compact in X. Here, $X\setminus C$ denotes setminus. Note that $V$ is an open neighborhood of $\{\infty \}$ , and thus, any open cover of $\{\infty \}$ will contain all except a compact subset $C$ of $X^{*}$ , implying that $X^{*}$ is compact.

That is, a topology has to be an open cover. This wording makes it clear where the open sets are. 67.198.37.16 (talk) 19:52, 29 August 2016 (UTC)[reply]

OK, I decided to "be bold" and just make this change. If you don't like it, then just revert. Its not that big a deal. 67.198.37.16 (talk) 20:13, 29 August 2016 (UTC)[reply]

Compactification of function spaces

There is certain relation to the space of continuous functions – namely, Alexandroff extension for $Ω$ corresponds to extension for the $C 0 (Ω)$ algebra with the (new) 1 element; nothing at all about compactifications of function spaces. But the IP got it all wrong. First of all, $C (X, ℝ)$ always contains a constant function $f \equiv 1$ (it’s namely $C 0$ which does not necessarily). Second, all these function spaces may not be compact unless $X$ is empty. Nuke this excretion of a tongue-tied student, Wikipedians. Incnis Mrsi (talk) 18:27, 12 April 2019 (UTC)[reply]

disadvantage

I believe that the following sentence is rather misleading.

the disadvantage (of the Alexandroff compactification) lies in the fact that it only gives a Hausdorff compactification on the class of locally compact, noncompact Hausdorff spaces

In a sense, this is a disadvantage... but we can do no better. Only Tychonoff spaces can be embedded into compact Hausdorff spaces, and if we want the embedding to be open we do need local compactness. To this effect, the Stone–Čech compactification does no better, we simply obtain Hausdorff by renouncing to getting an embedding.

It seems to me that the Alexandroff compactification is simply different from the Stone–Čech compactification; the universal property of the latter is useful, but the minimality of the former is useful, as well. Paolo Lipparini (talk) 21:28, 31 July 2021 (UTC)[reply]