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Jun 11, 2020 at 8:22 comment added Alok Many thanks for this!
Jun 9, 2020 at 14:54 comment added Nik Weaver The maximal ideals of $l^\infty$ correspond to ultrafilters on $\mathbb{N}$. The algebra $l^\infty(l^\infty)$ is isomorphic to $l^\infty(\mathbb{N}^2)$. Is every ultrafilter on $\mathbb{N}^2$ a product of two ultrafilters on $\mathbb{N}$?
Jun 9, 2020 at 14:01 comment added Alok Does this show that it is strictly bigger? Can't see it yet... could you give an outline please?
Jun 9, 2020 at 13:47 comment added Matthew Daws What happens if $A=\ell^\infty$?
Jun 9, 2020 at 12:25 review First posts
Jun 9, 2020 at 12:51
Jun 9, 2020 at 12:20 history asked Alok CC BY-SA 4.0