Espaces de suites réelles complètement métrisables

Volume 168 / 2001

Pierre Casevitz Fundamenta Mathematicae 168 (2001), 199-235 MSC: Primary 06F20, 46A19, 46A45, 54D55; Secondary 03E15, 54E52. DOI: 10.4064/fm168-3-1


Let $X$ be an hereditary subspace of the Polish space ${\mathbb R}^\omega $ of real sequences, i.e. a subspace such that $[x=(x_n)_n\in X$ and $\forall n$, $|y_n|\leq |x_n|]\Rightarrow y=(y_n)_n\in X$. Does $X$ admit a complete metric compatible with its vector structure? We have two results:

$\bullet $ If such an $X$ has a complete metric $\delta $, there exists a unique pair $(E,F)$ of hereditary subspaces with $E\subseteq X\subseteq F$, $(E,\delta )$ complete separable, and $F$ complete maximal in a strong sense. On $E$ and $F$, the metrics have a simple form, and the spaces $E$ are Borel (${\bf \Pi }_3^0$ or ${\bf \Sigma }_2^0$) in ${\mathbb R}^\omega $. In particular, if $X$ is separable, then $X=E$.

$\bullet $ If $X$ is an hereditary space, analytic as a subset of ${\mathbb R}^\omega $, we can find a subspace of $X$ strongly isomorphic to the space $c_{00}$ of finite sequences, or we can find a pair $(E,F)$ and a metric with the same properties around $X$. If $X$ is ${\bf \Sigma }_3^0$ in ${\mathbb R}^\omega $, we get a complete trichotomy describing the possible topologies of $X$, which makes precise a result of [C], but for general $X$'s, there are examples of various situations.


  • Pierre CasevitzSDAD – Université de Caen
    Campus II, Boulevard Maréchal Juin
    1, Esplanade de la Libération
    BP 5186
    F-14032 Caen Cedex, France

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