Talk:Lebesgue covering dimension

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The definition given uses the term "refinement", but incorrectly. A refinement is not a subcover, the elements of the refinement need only be open subsets of the original cover. I don't know if the definition of LCD is correct with "refinement" or "subcover", though. 4.31.92.84 05:15, 17 Dec 2004 (UTC)

As one of the authors, I'll comment that the page probably needs checking anyway. It is partly based on what I recall of the proof of the simplicial approximation theorem. Charles Matthews 14:10, 17 Dec 2004 (UTC)

FWIW, here's one problem with the definition as it currently stands: consider the unit disk in 2D. It can be covered by a slightly larger open disk. Since each point occurs only once, then can we conclude that the unit disk has topo dimension 1? or even 0? I don't think so...

I'm reading something old, maybe out-of-date; the definition is by Menger not Lebesgue. The space whose dimension we are trying to find out must be a metric space, the covering sets are closed not open, and the subcover must be finite (not just countable). Furthermore one must talk about embedding an n-dimensional topo set in a 2n+1 euclidean space. So for an n=1 curve, embed it in R^3. This curve can be covered with a finite number of closed sets in R^3, of arbitrarily small radius. The key to the proof seems to be "of arbitrarily small radius", since a 2D disk cannot be covered in this way, and still have every point belong to at most n+1=2 sets of the cover. Only a line satisfies this. I'm not sure why R^3 is needed,and why R^2 doesn't suffice for the proof. I'm also not clear if the closed vs. open distinction is important.

FWIW, the Menger sponge has (Menger) topological dimension 1, and every topo-dimension 1 curve is homeomorphic to the Menger sponge. I have no idea if the Lebesgue topo dimension would assign a different value to the Menger sponge. linas 14:38, 19 May 2005 (UTC)

Never mind. The key to lebesgue defnition is that every (finite?) open cover must have a (finite?) refinement where each point is contained in no more than n+1 sets of the refinement. The disk is 2D, because it has open covers that cannot be refined so that each point is only in 2 sets; 3 is the minimum. The Menger sponge still seems to be 1D. I don't know if the finite/infinite distinction might change this. linas 15:17, 19 May 2005 (UTC)
I think should be every open cover (not necesarily finite) with a locally finite refinement... If there exist open covers without locally finite refinements the top. dim. would be infinite.

Hmm. The edit by User:Kuratowski's Ghost seems really really wrong to me. The original wording, refinment was correct, subcover is wrong. Here's a counter-example: Take the unit disk in 2D and slice it up into 12 pie slices. Grow each pie slice a bit, so each pie slice overlaps. That's an open covering of the unit disk by 12 open sets. Points near the center of the disk now belong to 12 open sets, and its impossible to discard any of the twelve and still have a cover. This implies that the 2D disk is 11-dimensional, clearly false. So that definition is broken. By contrast, those 12 slices can be refined so that the points in the center are covered by no more than three open sets. So the earlier version of the article was correct. linas 03:29, 20 May 2005 (UTC)

Contents

[edit] Construction of the Sierpinski Carpet

I'm planning on adding the following text to the main article real soon now. Please review.

The definition of the Lebesgue covering dimension can be used to build some unusual toplogical sets, such as the Sierpinski carpet. A construction can proceed as follows.

Consider, for example, a finite open covering for the two-dimensional unit disk. This covering can always be refined so that no point in the disk belongs to more than three sets. Fixing this covering, remove all of the points in the disk that belong to three sets. Depending on the refinement, this will leave possibly one or more holes in the disk. The remaining object is again two-dimensional, and again has a finite open cover. The process of selecting a cover and refining, and then punching out holes can be repeated, ad infinitum. The resulting object is homeomorphic to the Sierpinski carpet. What is curious about this construction is that the carpet has a Lebesgue covering dimension of one, and not two. Given any open covering of the carpet, one can always find a refinement such that every point belongs to at most two sets. The proof of this is essentially by contradiction: were there a covering which required membership to three sets, then the affected area would have been punched out during the construction phase. As open covers are at most countable, every such case is handled during construction. Similar constructions can be performed in higher dimensions; the three-dimensional analogue is called the Menger sponge. Curiously, the Lebesgue covering dimension of the Menger sponge is again one.


This construction of the Sierpinski carpet is at best incomplete because it assumes that only a countable number of different covers are possible and that every cover of the sponge extends to a cover of some step in the construction. Neither of these facts are evident. Grubb257 17:57, 27 September 2007 (UTC)

I think this section needs to be rewritten. Point to the article on the Sierpinski carpet and state that it has covering dimension equal to one and is a universal curve. However, the construction in this article is unfixable as far as I can see.131.156.3.80 20:04, 28 September 2007 (UTC)

The Menger sponge has some additional curious properties. It is the universal curve. By this we mean that any possible one-dimensional curve (embedded in any number of dimensions) is homeomorphic to a subset of the Menger sponge. In a more restricted sense, any possible one-dimensional object embedded in the two-dimensional plane is homemomorphic to a subset of the Sierpinski carpet. Note that by curve we mean any object of Lebesgue dimension one; this includes trees and graphs with an arbitrary (countable) number of closed loops.

linas 03:51, 20 May 2005 (UTC)

[edit] Refinement or Subcover?

An earlier version of the article spoke of a refinements instead of subcovers, but the definition of refinement that was given was that of a subcover, so I changed it to subcover. But this indeed doesn't make sense as the examples show, I assume it really meant refinement but explained it incorrectly, will change again. Kuratowski's Ghost 09:40, 20 May 2005 (UTC)

The correct definition of the Lebesgue covering dimension uses refinements. As of today, the article is correct.Grubb257 17:58, 27 September 2007 (UTC)

[edit] why only finite dimension

Is there any particular reason why this definition restricts itself to finite dimensional cases? Can this definition not be used to distinguish between spaces whose dimensions are different infinite cardinals? Lethe | Talk 07:14, July 11, 2005 (UTC)-07:14, July 11, 2005 (UTC)

I don't know. One might guess that it would, but math often has paradoxes lurking betwixt the non-integer cardinals. linas 15:07, 11 July 2005 (UTC)
That sounds scary. But I'm not convinced. For example, I've seen lots of linear algebra books which deal only with finite dimensional vector spaces, solely as a matter of convenience. One can consider vector spaces whose dimension is any cardinality, and while certain theorems no longer hold (e.g. vector space isomorphic to its dual), certainly no paradoxes arise. The authors who do it are just being lazy (or are shielding the reader from unnecessary complication). -Lethe | Talk 17:14, July 11, 2005 (UTC)
Maybe one persons paradox is another person's complication. The set of basis functions for an infinite-dimensional Hilbert space is weakly convergent to zero. Maybe not a paradox, but certainly counter-intuitive; how can a series of vectors, all of whose lengths are one, converge to zero? Whatever notion you may have of Cauchy sequences in topology, infinite-dimensional vector spaces have a zoo of counter-intuitive examples. The topology of infinite dimensional spaces is particularly bizarre, and requires considerable care. Take for example the space of functions that are smooth (differentiable) and also integrable. Sounds easy, but its just crazy enough that considerable care was needed to correctly construct a self consistent definition -- which is the Sobolev space, and is a non-trival topic. The glossed-over definitions given in calculus books are neither supersets nor subsets of the Sobolev space.
Seriously, oceans of bizarre math occur at that first cardinal. Lets see Banach-Tarski paradox, the closure of the rationals by the p-adic numbers, fractals such as cantor set and sierpinski gasket ... seriously .. the sierpinski gasket looks sooooo 2D and yet its not... I find that paradoxial in and of itself. Surely you are trying to get a rise out of me? linas 02:34, 12 July 2005 (UTC)
I just want to use this definition for some infinite dimensional topological spaces. I know that funny things can happen in infinite dimensional spaces, but I don't think that's anything to be afraid of. -Lethe | Talk 18:45, July 12, 2005 (UTC)
Oh, well, sure I guess so, ... I know nothing about infinite dimensional topological spaces right now. If this is how other mathematicians define things, then sure. By experience, whenever I get to infinity, I make some blunder; I've been bitten many a time, and I'm not over it. Wasn't trying to say anything deeper than that. linas 05:13, 13 July 2005 (UTC)
There is no reason why the minumum number n... should exist, in this case the topological dimension would be infinite. This should be true for vector spaces of infinite dimension. maybe this should be written here more explicitly.

—The preceding unsigned comment was added by 150.214.102.129 (talk) 12:12, 9 May 2007 (UTC).

[edit] Menger sponge universal property

Am I right in thinking that "any object of Lebesgue dimension one", means compact metrizable spaces of Lesbesgue dimension one. Certainly object can't mean topological space or we could take a discrete space with too big a cardinality as a counterexample. I'm not CERTAIN we need metrizable (what is the dimension of {0,1}^(big cardinal)?) I'd actually really like to know. A Geek Tragedy 14:59, 11 February 2007 (UTC)

The dimension of {0,1}^(big cardinal) is still 0. Grubb257 18:00, 27 September 2007 (UTC)