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)

Okay, thanks for clearing this up. As I understand it now a TVS has a uniform structure and thus it makes sense to speak about complete TVSes. A Banach space is a CTVS for a norm-induced uniform structure, a Frechet space is a CTVS for a metric-induced uniform structure. This is more general because a norm implies a metric, but not vice versa. A metric does not imply a norm, bacause of this property of a norm p: p(a v) = |a| p(v). --MarSch 12:55, 5 Jun 2005 (UTC)

Assuming my above comment is correct, I wonder if the term Metric_vector_space used. Then a Banach space would be a complete normed vector space (CNVS) and a Frechet space would be a complete metric vector space (CMVS).--MarSch 13:01, 5 Jun 2005 (UTC)

Your understanding (comment) is correct. As for the term Metric_vector_space, I have not seen this usage before, but perhaps the following explanation will help.

I have a mathematics background, thus I am primarily interested in topological and algebraic structures. A topological vectors space is a combination of a linear structure and a uniform structure (so perhaps it should better be called uniform vector space). Generally a uniform structure can be described by a family of pseudometrics (see gauge space). If the family consists of only one pseudometric we get a pseudometric space. The topology of a locally convex space (nearly all interesting topological vector spaces are locally convex) can be described by a family of seminorms, which makes this a special case of gauge space. If the family is countable, the seminorms can be used to define a translation invariant metric. In other words a locally convex space is metrisable (even by a translation invariant metric) if and only if its topology can be described a countable family of seminorms. If the family of seminorms consists of just a single seminorm, then the locally convex space is a seminormable space.

What I am trying to say is norms and metrics are just ways to specify a topology, they are convenient, but generally we are interested in the topological structure (usually there are several different families of semi norms generating the same topology). Physicists often use normed vector spaces, so they talk about normed vector space, which in my opinion obscures the important part, namely that the topology is normable. A metrisable vector space is quite abstract and probably not used as often in physics as normed vector spaces therefore the term Metric_vector_space is not widely used. I prefer to talk about normable topology and metrisable topology.

MathMartin 13:44, 5 Jun 2005 (UTC)

thanks, that is most enlightening. Now we only need to get this into the article --MarSch 16:30, 5 Jun 2005 (UTC)

Frechet product metric

I tripped accross the following definition of a Frechet product metric

$d(x,y)=\sum _{n=0}^{\infty }{\frac {1}{2^{n}}}{\frac {\vert x_{n}-y_{n}\vert }{1+\vert x_{n}-y_{n}\vert }}$

as a metric on the product topology of $\mathbb {R} ^{\omega }$, the product of a countable number of copies of the reals. This metric turns $\mathbb {R} ^{\omega }$ into a Frechet space. I guess I should add this as an example to the article. Not sure why I thought to mention it here first. linas 14:46, 21 November 2005 (UTC)

Another example

From Counterexamples in Topology:

Every separable metric space is homeomorphic to a subspace of Frechet space. A mapping is given by the function

$f(x)=\langle d(x,a_{n})\rangle$

where x is a point in the metric space, and the $\{a_{n}\}$ is a countable dense subset of the metric space. The angle brackets are used to denote a point in Frechet space as $y=\langle y_{n}\rangle$.

I suppose I should add the above to the article, but I thought I'd mention it here first. linas 15:23, 21 November 2005 (UTC)

I don't understand what the angle brackets mean, nor can I understand your explanation. -lethe ^talk 15:55, 21 November 2005 (UTC)

I looked in Counterexamples, and they apparently use the angle brackets to denote ordered families, as in products. So OK. -lethe ^talk 16:40, 21 November 2005 (UTC)

I'm not sure if that statement stated in 2005 is really that impressive, since I have heard that every metric space (not only the separable ones) is isometrically isomorphic (not only homeomorphic) to a subset of a Banach space (not only a Frechet space). Unfortunately this is nothing which has to do something with "Frechet Spaces" anymore...

If you are interested: If (X,d) is a bounded metric space, the isometry can be constructed very easily: Just take C_b(X)={f:X to R : f continuous and bounded} as the Banach space with the sup-Norm and map every p to the function: $a\mapsto d(a,p)$.

If X is not bounded, you have to be a bit more careful and dualize twice: Let C_L(X) be the set of all f\in C_b(X) which are 1-Lipschitz, this is a subset of a Banach space and therefore a metric space, now we construct the space E:= C_b( C_L(X) ) of all continuous bounded functions defined on that new space, and every p in X is mapped onto its evaluation map, i.e. p \in X is mapped onto the function that evaluates a function of C_L(X) at point p. This becomes an isometry.

Sorry, if nobody wanted to read all this ;) --131.234.106.197 (talk) 17:23, 15 April 2008 (UTC)