Weak star separable and separable quotient problem - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T17:39:40Z http://mathoverflow.net/feeds/question/79073 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/79073/weak-star-separable-and-separable-quotient-problem Weak star separable and separable quotient problem Qingping Zeng 2011-10-25T13:02:22Z 2011-10-25T15:06:45Z <p>My first question is the following: </p> <p>Q1: Let $X$ be a Banach space. If its dual $X^*$ is weak* separable, does $X$ admit an infinite-dimensional and separable quotient $X/M$? </p> <p>To the best of my knowledge, the dual $X^*$ is weak* separable, when $X$ satisfies one of the following:</p> <p>(i) $X$ is separable;</p> <p>(ii) $X$ is the dual of a separable Banach space;</p> <p>(iii) $X$ is Hereditary Indecomposable Banach space. That is, every infinite-dimensional closed subspace of $X$ can not be written as a direct sum of two infinite-dimensional closed subspaces. </p> <p>And, I see that if $X$ satisfies (i) or (ii), $X$ admit an infinite-dimensional and separable quotient. </p> <p>Q2: Is it true that $X$ admit an infinite-dimensional and separable quotient, if $X$ satisfies (iii) ? </p> http://mathoverflow.net/questions/79073/weak-star-separable-and-separable-quotient-problem/79088#79088 Answer by Bill Johnson for Weak star separable and separable quotient problem Bill Johnson 2011-10-25T15:06:45Z 2011-10-25T15:06:45Z <p>Q1 is equivalent to the separable quotient problem. Indeed, given $Y$ infinite dimensional, let $W$ be any separable infinite dimensional subspace of <code>$X^*$</code> and let $Y$ be the annihilator of $W$ in $X$. Then the dual of $X/Y$ is the weak<code>$^*$</code> closure of $W$ in <code>$X^*$</code>. </p>