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Oct 12, 2018 at 15:46 comment added YCor By the way, one does not need a basis, a linearly independent subset is enough, and such subsets of cardinal $2^{\aleph_0}$ can be explicitly constructed.
Oct 12, 2018 at 10:46 vote accept Dominic van der Zypen
Oct 12, 2018 at 10:02 history answered dan_fulea CC BY-SA 4.0