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Jul 30, 2013 at 15:29 vote accept Cameron Buie
Jul 19, 2013 at 13:17 comment added François G. Dorais Yes. The backward half of the equivalence, in the order you wrote them.
Jul 19, 2013 at 13:15 comment added Cameron Buie (+1) So, if I understand you correctly, Nagata-Smirnov and Bing are both equivalent (in ZF) to the statement "all metric spaces are paracompact"? What do you mean by "the backward implications"?
Jul 19, 2013 at 13:10 history answered François G. Dorais CC BY-SA 3.0