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

DOI:
https://doi.org/10.1090/S0002-9939-1984-0728367-6

MathSciNet review:
728367

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Abstract | References | Similar Articles | Additional Information

Abstract: Given an arbitrary infinite-dimensional separable complete linear metric space , there exists a direct sum decomposition such that each summand intersects every linearly independent Cantor set in (this decomposition can be considered as a linear analogue to the classical Bernstein's decomposition into totally imperfect sets).

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

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

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Additional Information

DOI:
https://doi.org/10.1090/S0002-9939-1984-0728367-6

Keywords:
Linear metric spaces,
homeomorphisms,
Bernstein's sets

Article copyright:
© Copyright 1984
American Mathematical Society