From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik) Subject: Re: Heine-Borel generalization? Date: 3 May 2000 14:03:25 -0400 Newsgroups: sci.math Summary: [missing] In article <8epje4$9hf$1@nnrp1.deja.com>, wrote: :Is there a generalization of the Heine-Borel theorem to other spaces :than R^n? : :Brian F. For normed linear spaces, this is as far as one can go. The "compact iff closed and bounded" property is confined to finite dimensional spaces. (In an infinite dimensional normed space, you can find an infinite set of unit vectors, with distance >1/2 between any two different ones -- such a set is closed and bounded but not compact; cover each vector with a ball of radius 1/8). (Dunford-Schwarz). A version of it ("there is an open neighborhood of 0 whose closure is compact iff the space is finite dimensional" can also be extended to other topological vector spaces. For metric spaces, boundedness is not a topological property at all; you can have a bounded space homeomorphic to an unbounded space. (Exercise: every metric space is homeomorphic to a bounded metric space.) An amendment is possible: replace boundedness by total boundedness, and closedness by completeness. (Note: Completeness is not a topological property either, but in this case it works.) Cheers, ZVK(Slavek). ============================================================================== From: ullrich@math.okstate.edu (David C. Ullrich) Subject: Re: Heine-Borel generalization? Date: Thu, 04 May 2000 14:49:32 GMT Newsgroups: sci.math On 3 May 2000 14:03:25 -0400, kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik) wrote: >In article <8epje4$9hf$1@nnrp1.deja.com>, > wrote: >:Is there a generalization of the Heine-Borel theorem to other spaces >:than R^n? >: >:Brian F. > [...] > >For metric spaces, boundedness is not a topological property at all; you >can have a bounded space homeomorphic to an unbounded space. Otoh if V is a topological vector space there is a standard notion of "bounded subset"; if V is a metrizable topological vector space then a bounded subset is not at all the same thing as a subset which is bounded as a subset of the corresponding metric space. A set E is bounded if for every neighborhood of the origin U there exists t > 0 such that E is contained in tU. That's not a purely topological property but it _is_ a purely topological- vector-space property, well-defined independent of whether the space is even metrizable. This is unfortunate because it can cause confusion - people sometimes think that a bounded subset of V is a subset of finite diameter in "the" metric. Otoh it's fortunate because it allows one to make sense of what I said yesterday in this thread (among other reasons.) >(Exercise: >every metric space is homeomorphic to a bounded metric space.) An >amendment is possible: replace boundedness by total boundedness, and >closedness by completeness. > >(Note: Completeness is not a topological property either, but in this case >it works.) > >Cheers, ZVK(Slavek). ============================================================================== From: ullrich@math.okstate.edu (David C. Ullrich) Subject: Re: Heine-Borel generalization? Date: Wed, 03 May 2000 18:56:00 GMT Newsgroups: sci.math On Wed, 03 May 2000 16:17:28 GMT, brian_flangsmythe@hotmail.com wrote: >Is there a generalization of the Heine-Borel theorem to other spaces >than R^n? Well, it's certainly false in "most" infinite-dimensional topological vector spaces (false in an infinite-dimensional space with a topology given by a norm, for example.) It's true in spaces with the so-called "Heine-Borel property" - I'll let you guess what the definition of that is. The only non-trivial example of a space with the Heine-Borel property that springs to _my_ mind would be the space of holomorphic functions in some open set. You give this the topology of uniform convergence on compact sets and then Montel's theorem says precisely that closed and bounded sets are compact. >Brian F. > > >Sent via Deja.com http://www.deja.com/ >Before you buy.