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Talk:Fréchet space
Complete???
what does it mean that a Frechet space is complete??? Excerpt fromcomplete space follows: "Note that completeness is a property of the metric and not of the topology, meaning that a complete metric space can be homeomorphic to a non-complete one. An example is given by the real numbers, which are complete but homeomorphic to the open interval (0,1), which is not complete. Another example is given by the irrational numbers, which are not complete as a subspace of the real numbers but are homeomorphic to NN (a special case of an example in Examples above)."MarSch 16:53, 18 Mar 2005 (UTC)
- There is no problem, as the metric is a given. A LCTVS space with a translation-invariant metric may not be complete, in which case it is not a Frechet space. Since any TVS is a uniform space with a translation-invariant uniform structure, and so there is a definition of completeness. Thus for TVSs completeness does not depend on having a metric, but it is quite common for there to be two inequivalent metrics, one of which gives a complete space and one of which does not. (e.g. the Banach space l1 with the metric from any other p-norm). The problem with the examples given is that they are not uniformly continuous (and Q is not homeomorphic to N, but that is another issue). HTH Andrew Kepert 22:57, 20 Mar 2005 (UTC)
- the def starts: "A topological vector X space is a Fréchet space iff it satisfies the following three properties:* it is complete". This apparently only makes sense after condition 3) states there is a metric. But if there is a metric then isn't it a normed VS? and thus Banach? How are Frechet spaces more general?MarSch 14:19, 23 Mar 2005 (UTC)
- I am not sure I understand you correctly. Anyway like Andrew Kepert said, completeness is defined in reference to a uniform structure. A metric induces a uniform structure but not vice versa, so we do not need a metric to talk about completeness. The second misunderstanding is although a norm induces a metric the converse is not true. Usually you cannot construct a norm using a metric. A Frechet space is a complete topological vector space (with a uniform structure) but unlike a Banach space the topology is not given by a norm. MathMartin 19:44, 9 Apr 2005 (UTC)
- the def starts: "A topological vector X space is a Fréchet space iff it satisfies the following three properties:* it is complete". This apparently only makes sense after condition 3) states there is a metric. But if there is a metric then isn't it a normed VS? and thus Banach? How are Frechet spaces more general?MarSch 14:19, 23 Mar 2005 (UTC)
Last updated: 05-27-2005 03:26:49
03-10-2013 05:06:04
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