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Jan 12, 2023 at 16:47 comment added Gerry Myerson Mathoverflow is not for discussions. It is for tightly targeted answers to well-posed questions.
Jan 9, 2023 at 12:58 comment added S Argyros If $X$ does not have $w^*$ separable dual then for every separable subspace $Y$ of $X$ there exists a non separable $Z$ such that $Y \cap Z = 0$. Indeed choose $( g_n )$ in $X^*$ such that the subspace $Z= ( g_n )_\bot$ has trivial intersection with $Y$. The subspace $Z$ is non separable since otherwise $X^*$ is $w^*$ separable.
Jan 8, 2023 at 18:26 comment added Bill Johnson In case it is not obvious to the two or three others following this thread, we should note that Theorem 3 above implies that the famous "separable quotient problem" (whether every infinite dimensional Banach space has a separable infinite dimensional quotient) is equivalent to the problem whether every separable subspace of every Banach space is quasi-complemented.
Jan 8, 2023 at 10:22 history edited S Argyros CC BY-SA 4.0
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Jan 6, 2023 at 17:43 comment added S Argyros I understand your point but I think that the discussion has a natural evolution. Anyone not interested on this he can simply skip it. It is so simple.
Jan 6, 2023 at 17:24 comment added Nik Weaver I appreciate your interest and zeal, but going on and on with these answers is probably not appropriate for this site...
Jan 6, 2023 at 17:15 history edited S Argyros CC BY-SA 4.0
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Jan 6, 2023 at 15:59 history edited S Argyros CC BY-SA 4.0
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Jan 6, 2023 at 15:42 history edited S Argyros CC BY-SA 4.0
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Jan 6, 2023 at 15:15 history edited S Argyros CC BY-SA 4.0
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Jan 6, 2023 at 15:00 history edited S Argyros CC BY-SA 4.0
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Jan 6, 2023 at 14:26 history edited S Argyros CC BY-SA 4.0
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Jan 6, 2023 at 14:18 history answered S Argyros CC BY-SA 4.0