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An infinite-dimensional pre-Hilbert space not homeomorphic to its own square

Author: Roman Pol
Journal: Proc. Amer. Math. Soc. 90 (1984), 450-454
MSC: Primary 57N20; Secondary 46C99, 54F45
MathSciNet review: 728367
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Abstract: Given an arbitrary infinite-dimensional separable complete linear metric space $ X$, there exists a direct sum decomposition $ X = {V_0} \oplus {V_1}$ such that each summand $ {V_i}$ intersects every linearly independent Cantor set in $ X$ (this decomposition can be considered as a linear analogue to the classical Bernstein's decomposition into totally imperfect sets).

Theorem. Each summand $ V$ of such a decomposition is not homeomorphic to its own square, and if $ T:V \to V$ is a linear bounded operator, then either the kernel or the range of $ T$ is finite-dimensional.

In the case of $ X = {l_2}$ this provides an example of a space $ V$ with the properties stated in the title, which answers a well-known question, cf. Arhangelskiĭ [A, Problem 21] and Geoghegan [G, Problem (LS 12)].

References [Enhancements On Off] (What's this?)

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Keywords: Linear metric spaces, homeomorphisms, Bernstein's sets
Article copyright: © Copyright 1984 American Mathematical Society

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