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Tsirelson's space was the first example of a Banach space which does not have a subspace isomoprhic to any of the classical spaces $\ell_p$, $1\leqslant p<\infty$, or $c_0$. As this space has a $1$-unconditional basis, it can also be viewed as a Banach lattice. Therefore Banach lattices need not contain isomorphic copy of any of the classical spaces $\ell_p$ or $c_0$.

What about in the case of a Banach lattice which is not discrete? Is there some positive result about the classical spaces which must be subspaces of a Banach lattice which is not discrete?

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  • $\begingroup$ Do you wish your non-discrete Banach lattice to contain the classical sequence spaces as closed sublattices, or just as closed subspaces? $\endgroup$
    – Jochen Glueck
    Commented Feb 23, 2018 at 22:23
  • $\begingroup$ By the way, two remarks concerning the spaces $c_0$ and $\ell^1$: (i) A Banach lattice $E$ is a KB-space if and only if $E$ does not contain $c_0$ as a closed subspace if and only if $E$ does not contain $c_0$ as a closed sublattice [1, Theorems 2.4.12 and 2.5.6]; (ii) A Banach lattice $E$ is reflexive if and only $E$ does not contain $c_0$ nor $\ell^1$ as a closed subspace if and only if $E$ does not contain $c_0$ nor $\ell^1$ as a closed sublattice [1, Theorem 2.4.15]. Reference: [1] "Peter Meyer-Nieberg: Banach Lattices (1991)" $\endgroup$
    – Jochen Glueck
    Commented Feb 23, 2018 at 22:28
  • $\begingroup$ I only care that it contains the classical spaces as subspaces. I do not require that they be sublattices. $\endgroup$
    – user114263
    Commented Feb 23, 2018 at 22:49
  • $\begingroup$ You can try to look at the paper Tokarev, E. V. A symmetric Banach function space that does not contain $\ell_p$ $(1\le p<\infty)$ and $c_0$ (Russian) Funktsional. Anal. i Prilozhen. 18 (1984), no. 2, 75-76 (it was translated into English). I should warn you that this author is known for making errors in proofs. $\endgroup$
    – Mikhail Ostrovskii
    Commented Feb 24, 2018 at 6:12
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    $\begingroup$ For an example illustrating @MikhailOstrovskii 's claim, see mathoverflow.net/questions/171680/… $\endgroup$
    – Tomasz Kania
    Commented Feb 24, 2018 at 10:58

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