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Paracompact subspaces in the box product topology

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Abstract

In 1975 E. K. van Douwen showed that if ( X n ) n ∈ ω ( X_n )_{ n \in \omega } is a family of Hausdorff spaces such that all finite subproducts ∏ n > m X n \prod _{ n > m } X_n are paracompact, then for each element x x of the box product ◻ n ∈ ω X n \square _{n \in \omega } X_n the σ \sigma -product σ ( x ) = { y ∈ ◻ n ∈ ω X n : { n ∈ ω : x ( n ) ≠ y ( n ) } is finite } \sigma ( x ) = \{ y \in \square _{n \in \omega } X_n : \{ n \in \omega : x (n) \neq y (n) \} \text { is finite} \} is paracompact. He asked whether this result remains true if one considers uncountable families of spaces. In this paper we prove in particular the following result: Theorem Let κ \kappa be an infinite cardinal number, and let ( X α ) α ∈ κ ( X_{\alpha } )_{\alpha \in \kappa } be a family of compact Hausdorff spaces. Let x ∈ ◻ = ◻ α ∈ κ X α x \in \square = \square _{\alpha \in \kappa } X_\alpha be a fixed point. Given a family R \mathcal {R} of open subsets of ◻ \square which covers σ ( x ) \sigma ( x ) , there exists an open locally finite in ◻ \square refinement S \mathcal {S} of R \mathcal {R} which covers σ ( x ) \sigma ( x ) . We also prove a slightly weaker version of this theorem for Hausdorff spaces with “all finite subproducts are paracompact" property. As a corollary we get an affirmative answer to van Douwen’s question.

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