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Oct 30, 2018 at 16:33 comment added Robert Furber @roger123 Every paracompact Hausdorff space admits a complete uniformity (see Kelley's General Topology Chapter 6, exercise L (d)), so a paracompact Hausdorff space is realcompact iff it has no discrete subset of (2-valued) measurable cardinality by Shirota's theorem. (See, for instance Gillman and Jerison's Rings of Continuous Functions Theorem 15.20). Therefore it is consistent that every paracompact Hausdorff space is realcompact.
Oct 30, 2018 at 15:18 history edited Todd Trimble CC BY-SA 4.0
Link (to a website that pirates copyrighted works according to a prominent MO user) removed
Nov 1, 2010 at 12:48 comment added roger123 Thank you. Paracompactness does not suffice here, right?
Oct 29, 2010 at 8:43 vote accept roger123
Oct 28, 2010 at 20:13 comment added David Roberts For reference: en.wikipedia.org/wiki/Realcompact_space
Oct 28, 2010 at 15:53 history answered Dmitri Pavlov CC BY-SA 2.5