Talk:Hilbert cube

Compact

The phrase: "Since l2 is not locally compact, no point has a compact neighbourhood, " seems a little overstretched. That a space is not locally compact could mean that all points but one have a compact neighborhood. 03:34, 8 November 2004 68.232.110.205

True. But since this sentence is merely of pedagogical nature anyway (mathematically, there is no reason that "one should expect" that compact subsets are finite dimensional), I would suggest to replace it by the following -- that is, unless one would simply delete it:

"Since no point of l2 has a compact neighbourhood, one might expect..."

(indeed l2 is homogeneous -- but I don't see the need to elaborate on this since this is just correcting a supposed misconception, "one might expect"...) 21:41, 13 January 2006 83.77.140.226

Contradiction?

I apologise, my math language is a bit rusty.... but I think some of the statements on this page don't add up, and I don't know enough about the subject to correct it.

"Topologically, the Hilbert cube may be defined as the product of countably infinitely many copies of the unit interval [0,1]. That is, it is the cube of countably infinite dimension."

"It's sometimes convenient to think of the Hilbert cube as a metric space, indeed as a specific subset of a Hilbert space with countably infinite dimension. For these purposes, it's best not to think of it as a product of copies of [0,1], but instead as

   [0,1] × [0,1/2] × [0,1/3] × ···;

for topological properties, this makes no difference."

and

"an open ball around p of any fixed radius e > 0 must go outside the cube in some dimension."

It seems to me that the third statement is dependent on the second, since if the first statement is true and truly does define a Hilbert cube then one simply needs to make the radius of the ball less than 1, then since in every dimension we are 'locally within' [0,1] any member of the ball must also be 'inside the cube in this dimension'.

Jheriko 13:48, 10 April 2006 (UTC)[reply]

Since nobody has corrected this I am going to do some research then replace/remove which ever part of this page is faulty.

Jheriko 10:34, 24 July 2006 (UTC)[reply]

Fixed. Jheriko 16:52, 20 September 2006 (UTC)[reply]

@Jheriko: "It seems to me that the third statement is dependent on the second, since if the first statement is true and truly does define a Hilbert cube then one simply needs to make the radius of the ball less than 1, then since in every dimension we are 'locally within' [0,1] any member of the ball must also be 'inside the cube in this dimension'." Your confusion is because of the subtleties involved in the definition of balls in the topologies concerned. The topology on a product of known topological spaces, called the product topology, is defined in such a way that such "balls/open sets" as you speak about, of a given size along every dimension, are not open. In fact the only open sets in product topology are those that contain the whole coordinate space along every "dimension" except finitely many. The first statement is true and doe indeed define a Hilbert cube, and the second statement gives a definition which is equivalent to this because of reasons which are expected infinite-dimensional analogues of the fact that [0,1] is homeomorphic to [0,1/2]. What the third statement means is that any open ball **wrt the whole space l2, which is neither a product, nor has the product topology from any master set**, of any small (non-zero) radius must go outside the cube in all except finitely many "dimensions". As such none of the three are faulty. For clarification you can add that the ball is wrt the whole space in a bracket. Nidhishunnikrishnan (talk) 13:40, 25 June 2017 (UTC)[reply]

The generators of the product topology (the basis of the topology) are called cylinder sets and all open sets are finite intersections/infinite unions of the cylinder sets. 67.198.37.16 (talk) 18:49, 11 September 2020 (UTC)[reply]

parallel edges

I wonder: the article states, "the length of each edge has a length 1/n ( $n\in \mathbb {N}$ ) with the extra condition that none of the edges are of equal length." But aren't any parallel edges of equal length?Orthografer 19:25, 26 August 2007 (UTC)[reply]

Reference '.X-OLAP' not understandable in English.

The second of the references (...X-OLAP) points to some IBM stuff presumably in Russian.

Perhaps the author of the reference can find a solution or put a warning.

--Uffe 08:19, 5 July 2006 (UTC)[reply]

Homeomorphic

"Topologically, the Hilbert cube is indistinguishable from..:"

is it homeomorphic? if so, let's just say so. Zero sharp 06:02, 4 August 2007 (UTC)[reply]

Misleading Language?

"The Hilbert cube is best defined as the topological product of the intervals [0, 1/n] for n = 1, 2, 3, 4, ... That is, it is a cuboid of countably infinite dimension, where the lengths of the edges in each orthogonal direction form the sequence ."

I believe I understand the intent of the language "cuboid of countable infinite dimension" but this language is somewhat misleading, as the Hamel dimension of the Hilbert cube as a Hilbert Space is decidedly NOT countable (as a consequence of the completeness of the Hilbert cube and the Baire Category Theorem). Perhaps this should be noted somewhere in the article? 71.61.180.25 (talk) 05:22, 28 January 2013 (UTC)[reply]

A serious article needed

This article is pathetic. Please, someone write it anew (I am just not up to the task). The article should mention the infinite dimensional results by Chapman, Szankowski, Toruńczyk, West, etc. Wlod (talk) 10:36, 28 December 2015 (UTC)[reply]

References

There are very few references in the section "Properties". I was trying to find some references but it seems to be difficult, especially for "Any infinite-dimensional convex compact in {\displaystyle l_{2}}l_{2} is homeomorphic to the Hilbert cube. " 213.225.6.170 (talk) 20:30, 18 February 2020 (UTC)[reply]