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Jun 21, 2013 at 20:28 vote accept Andrew Poelstra
Jun 21, 2013 at 20:28 comment added Andrew Poelstra I got it -- I was being dense, thanks!
Jun 21, 2013 at 20:22 comment added Andrew Poelstra Maybe I'm being dense, but why must the $A_n$'s be finite (or countable, even)?
Jun 21, 2013 at 18:48 comment added Simon Henry More generaly a product uniformity has a countable basis if and only if each term of the product has a countable basis and there is at most a countable number of term of the product which are non trivial. (where a uniforme strucure is trivial if it has no entourage other than $X \times X$ ).
Jun 21, 2013 at 18:45 history answered Simon Henry CC BY-SA 3.0