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Jul 21, 2010 at 7:49 comment added Pietro Majer note: the existence of a complementary space to $\mathbb{Q}$ requires the axiom of choice, but as François G. Dorais remarks, you just need an uncountable rational subspace with$ D\cap \mathbb{Q}=(0)$, that can be exhibited explicitly.
Jul 19, 2010 at 21:39 comment added Arin Chaudhuri Yes. This is quite clear.
Jul 19, 2010 at 21:36 comment added Pietro Majer expanded enough?
Jul 19, 2010 at 21:35 history edited Pietro Majer CC BY-SA 2.5
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Jul 19, 2010 at 21:32 comment added Arin Chaudhuri I understood it Daniel's comment, thanks.
Jul 19, 2010 at 21:31 vote accept Arin Chaudhuri
Jul 19, 2010 at 21:29 comment added Arin Chaudhuri I am sorry, I didn't get you, could you expand a bit?
Jul 19, 2010 at 21:26 history answered Pietro Majer CC BY-SA 2.5