I do not know if anyone has asked about this before: Does anyone know an example of an unbounded subset of a Frechet space which have finite diameter? I saw this in Conway (A course in functional analysis, page 107, execrise 4), in which he suggests every bounded subset of a Frechet has finite diameter(easy to prove), but he contends the opposite is not true. Is there some trivial examples? I just cannot imagine how this might happen judging from the geometrical picture of a TVS.
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The diameter is defined in terms of a metric defining the topology? – Daniel Fischer Jul 18 '13 at 02:33
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Yeah, Conway did not define it, but I suppose $diam(M)=\sup d(a,b):a,b\in M$ as usual. – Bombyx mori Jul 18 '13 at 02:34
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If the diameter is defined in terms of a metric $d$ defining the topology, then any ball
$$B_\varepsilon(0) = \{x \in E\colon d(x,\,0) < \varepsilon\}$$
in a Fréchet space $E$ that is not normable is an example of a set with finite diameter that is not bounded.
$B_\varepsilon(0)$ is a neighbourhood of $0$, and a space is normable if and only if it is Hausdorff, locally convex, and locally bounded, the latter meaning it has a bounded neighbourhood of $0$.
Examples of Fréchet spaces that are not normable are $C(U)$ for $U \subset \mathbb{R}^n$ open (and nonempty) with the topology of locally uniform convergence, $\mathcal{O}(U)$, the space of holomorphic functions on an open subset $\varnothing \neq U \subset \mathbb{C}^n$, also with the topology of locally uniform convergence, and $C^\infty(U)$ with the topology of locally uniform convergence of all derivatives. There are many more.
Daniel Fischer
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