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How can I show that, if a Frechet space is not normable then, there is no countable base of bounded sets.

A collection $\Gamma$ of subsets of X is called a base for bounded sets, if for any bounded set C there is a $B_{0}$ such that C is contained in $B_{0}$

Mahmut çükübikyan

user29253
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  • Do you assume that elements of base are open sets? – Norbert May 14 '12 at 17:17
  • No just sets in $X$. – user29253 May 14 '12 at 18:39
  • In the definition of "base of bounded sets" sets themselves should also be bounded. So "base of bounded sets" means that A collection Γ of bounded subsets of X is called a base for bounded sets, if for any bounded set C there is a B0 such that C is contained in B0 –  May 14 '12 at 22:50
  • Of course why should we call as a base of bouded sets? – user29253 May 16 '12 at 05:49

1 Answers1

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If $C$ is a subset of $X$, then $C$ is bounded if and only if the absorbent hull of $C$ in $X$ is bounded. Now given an absorbent bounded subset $C$ of $X$, we may define a norm on its span so that $C$ is the unit ball. In short, we may find a continuous map $W \to X$ where $W$ is a normed space and $C$ is the image of the unit ball in $W$. Conversely, the image under any continuous linear map from a normed space to $X$ of the unit ball in the domain is a bounded subset of $X$.

If $C$ is furthermore assumed closed, then $W$ will be a Banach space. Since the closure of a bounded set is again bounded, we may restrict attention to closed bounded sets. The existence of a countable basis of (closed) bounded sets in $X$ would then be equivalent to the existence of a collection of Banach spaces $W_n$ with continuous injections into $X$, such that any injection of a Banach space $W$ into $X$ factors through some $W_n$.

Now any Frechet space is ultrabornological, i.e. is the locally convex inductive limit of Banach spaces. Thus the existence of a countable basis of closed bounded subsets in $X$ would imply that $W$ is the locally convex inductive limit of a sequence of Banach spaces. A standard argument using the Baire category theorem would then imply that $W$ actually is a Banach space.

There are variants on the above argument, and one can probably make it more direct. One variant is as follows: if $X$ admits a countable basis of bounded sets, then the strong dual $X'_b$ admits a countable basis of seminorms, and so is again a Frechet space. But if the strong dual of a Frechet space is metrisable, then the original Frechet space is actually normable, and so is a Banach space. (If you unwind it, the proof of this latter statement is similar to the argument given above.)

Matt E
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