Is there a Banach space X with the property that neither X nor its dual X' has a boundedly complete basic sequence?
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$\begingroup$ Jog my memory: Is there known to exist a space $X$ such that its dual $X^*$ does not contain a boundedly complete basic sequence? If so, $X^*$ must be HI, but $X^*$ being HI is not sufficient. $\endgroup$– Bill JohnsonCommented Oct 12 at 18:29
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1$\begingroup$ @Bill Johnson: Must be HI or it does not contain an unconditional basic sequence? i.e. it is HI saturated? $\endgroup$– S ArgyrosCommented Oct 13 at 10:21
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$\begingroup$ I think that for $X$ separable such that $X^*$ does not contain boundedly complete basic sequence then $X^*$ is non separable. In my view this can be proved with the use of Ramsey theory. I mean either the classical infinite Ramsey Theory or Gowers Dichotomy. $\endgroup$– S ArgyrosCommented Oct 13 at 11:37
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1$\begingroup$ Thank you both for your help. The property I'm trying to use is that either X or X* has a boundedly complete basic sequence, and I was wondering how general it is. I couldn't find any examples that didn't have that property. $\endgroup$– C StuartCommented Oct 14 at 13:46
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1$\begingroup$ @C Stuart: This is an interesting question. I believe that it has a negative answer. The candidate space is related to Gowers tree space and it is not explained in a mathoverflow answer.I can give you some more details if you send me an email. $\endgroup$– S ArgyrosCommented Oct 14 at 14:09
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