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How can we show that $c_0$ has no closed complement in $l^\infty$. Similarly $C([0,1])$ has no closed complement in $B([0,1])$

flag	asked Dec 13 at 2:13 Prasanth 11●1

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To put it a little bluntly: what's your motivation for asking? (see mathoverflow.net/howtoask#motivation) – David Roberts Dec 13 at 3:04

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$c_0$ is complemented in any separable superspace, David. It is natural to ask if it is complemented in any superspace, which is equivalent to the first question because $\ell_\infty$ is injective. – Bill Johnson Dec 13 at 3:59

@Bill That's a nice way to look at it :) – David Roberts Dec 13 at 4:05

Agree with @David. It's not for answerers to provide motivation! – Anthony Quas Dec 13 at 6:18

Bill Johnson · Answer 1 · 2012-12-13 04:02:52Z UTC

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For the first question, see Theorem 2.5.5 in the book of Albiac and Kalton. The second question is immediate from the first and the easy fact that $C[0,1]$ has a complemented subspace isometric to $c_0$.

answered Dec 13 at 4:02

Bill Johnson
14.4k●1●33●60

closed complement

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