Talk:Inductive dimension

Please help with this page

OK, this is a topic that was clearly missing, but unfortunately I don't have any texts on the subject and couldn't find much online. Much of this is based on a post to sci.math by Ilias Kastanis, who's usually pretty reliable, but surely someone can do better than a Usenet post.

So it needs checking, and references, and explanation of the importance of the concept and the relationships among the various dimensions. Also examples of when the dimensions are different could be nice. Hoping someone can help out! --Trovatore 07:08, 16 November 2005 (UTC)[reply]

The definitions given look like Menger's (Japanese encylopedia article dimension theory). Minimum hypotheses for a sensible theory are something like normal space. Charles Matthews 17:39, 17 November 2005 (UTC)[reply]

Inequality

One of the sources I looked at claimed that it's trivial that ind(X)≤Ind(X). I see that, provided that singletons are closed (that is, that X is a T1 space). I don't see how to do it without any separation axiom at all.

That seems right. The reference I added explicitly assumes T1.

Does the result hold without any separation axiom?

Dunno

Are ind(X) and Ind(X) currently correctly defined, in the case that X is not assumed to satisfy any separation axiom?

Nearly; I'll make changes; please review.

--Trovatore 19:52, 16 November 2005 (UTC)[reply]

linas 02:16, 18 November 2005 (UTC)[reply]

As mentioned on my talk page, I do have the book reference, and I'm sitting down to read it now. I'll check/review/add any simple, succinct results as I come across them. BTW, I thought it was Urysohn too, but that book seems to spell it as Uryson. Cyrillic spellings ... linas 03:25, 17 November 2005 (UTC)[reply]

Cool; thanks. As to the spelling, we have an article on Paul Samuilovich Urysohn, so maybe it would be good to keep that spelling (also it's the one I've seen most often). Of course it's not surprising that there'll be different Romanizations of the name. --Trovatore 04:21, 17 November 2005 (UTC)[reply]

(Actually, shouldn't it be "Pavel", at least? Romanization is one thing, Anglicization is another. I get very annoyed with people who talk about "George Cantor".) --Trovatore 04:22, 17 November 2005 (UTC)[reply]

Probably. I know the cyrillic alphabet and a half-dozen words of slavic; that's all. linas 02:16, 18 November 2005 (UTC)[reply]

Definition of small ind

Today, for some reason, I don't feel like editing the article. Let me instead propose a revised version here, as below:

Given a point x∈X, one says that $\operatorname {ind} _{x}(X)\leq n$ if for each open neighborhood U of x there is an open subset V of U, also containing x, and whose closure is contained in U, such that the boundary of V has small inductive dimension less than or equal to n-1. If the above holds for a given value n but not for n-1, then one says $\operatorname {ind} _{x}(X)=n$ . The small inductive dimension of X is then defined as

\operatorname {ind} (X)=\sup\{\operatorname {ind} _{x}(X):x\in X\}

Notice two basic changes compared to current article:

The closure of V must be contained in U
The "smallest" is replaced by "sup" of the smallest.

So the first point is critical; it at least potentially gives you a different value (not, certainly, that I have an example in mind). I've edited the article accordingly. The second one seems to be more a style point (the least n that works for every x is the sup of the ones that work for a particular x). My initial reaction is I'm not so excited about the second change, but I wouldn't sharply oppose it either, if someone had a strong preference for it.

Note that we have not explicitly dealt with the infinite-dimensional case; using the sup version or the every-x version wouldn't change the answer there (if I haven't missed something) but it might change the exposition a little. --Trovatore 04:00, 18 November 2005 (UTC)[reply]

OK. I somehow missed/misread "the smallest n for every x" as "out of all the x, pick the smallest n". I humbly suggest the current wording might be slightly ambiguous, but I am too tired to propose anything right now. linas 04:39, 18 November 2005 (UTC)[reply]

Hm, I don't think it's a real ambiguity, but certainly it's better if things are hard to misread. Somehow the "sup" thing still strikes me as an unneccessary extra layer, though. But I don't feel strongly about it; I can see that the sup approach has advantages as well. --Trovatore 06:35, 18 November 2005 (UTC)[reply]

If this looks good to you, we should copy it into the article.

The Fedorchuk reference I gave also has another, alternate definition which is somewhat prettier but more verbose. I'll type it in if encouraged.

It also defines a different type of ind, called ind^p, based on being able to separate a pair of points, and then presents various theorems and inequalities relating it to ind. It argues that in a way, ind^p is the better partner to big Ind as its definition is more symmetric. linas 03:13, 18 November 2005 (UTC)[reply]

Definition of big Ind

The only difference that I see between this article and Fedorchuk for big Ind is that Fedorchuk requires that the closure of V be contained in U.

Curiously, the definition of the big Ind isn't given by the "sup of the smallest", although it seems "obvious" that a smart reader should just "assume" this. I am a newbie to topology; I'm just noting what the book states. linas 03:31, 18 November 2005 (UTC)[reply]

Reference for infinite-dimensional spaces

The Fedorchuk chapter of this book: Arkhangelskii, A.V.; Pontryagin, L.S. (1990). General Topology. Vol. I. Berlin, DE: Springer-Verlag. ISBN 3-540-18178-4. has a chapter on transfinite (i.e. bot countable and uncountable) dimensional spaces. 67.198.37.16 (talk) 23:24, 27 November 2023 (UTC)[reply]

So I don't have that book, but I would be surprised to see inductive dimension extended into the transfinite. What do you do at limit ordinals? What is the boundary of an $(\omega ^{2}+\omega \cdot 17+47)$ -dimensional ball? Lebesgue covering dimension seems more tractable but I haven't really thought about it much. --Trovatore (talk) 00:18, 28 November 2023 (UTC) Never mind. It's not defined in terms of Euclidean balls per se. The existing definition might work with a very minor tweak (replace "dimension less than or equal to n−1" by "dimension less than n"). But again I haven't thought much about it. --Trovatore (talk) 00:35, 28 November 2023 (UTC)[reply]