My question is inspired from the concept of super-reflexivity which was defined by James here: https://www.cambridge.org/core/journals/canadian-journal-of-mathematics/article/superreflexive-banach-spaces/F295D854464196BA440C80E4F4E22BF6. Here super-reflexivity is defined in terms of finitely representability. A Banach space $B$ is super-reflexive if no non-reflexive Banach space is finitely representable in $B$. Now if I replace "finitely-represetable" by "embeds" I come to the question, is there a Banach space $B$ such that no non-reflexive Banach space embeds into $B$? And a related question, can a non-reflexive space embed into a reflexive space?
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4$\begingroup$ If embedding means "isomorphic embedding'', then a space that embeds into a reflexive space is reflexive itself. If embedding means "continuous injection'', look at the example of $\ell_1$ and $\ell_2$. $\endgroup$– Dirk WernerCommented Apr 7, 2022 at 19:32
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$\begingroup$ If you work in a set theory where the Hahn-Banach theorem is not a theorem, then it is possible for a non-reflexive space Banach space to be a closed subspace of a Banach space. For instance, in Zermelo-Fraenkel without choice, if every finitely-additive probability measure on $(\mathbb{N},\mathcal{P}(\mathbb{N}))$ is countably additive then $\ell^\infty$ is reflexive (its dual space is $\ell^1$). But $c_0 \subseteq \ell^\infty$ is not reflexive, because $c_0^{**} \cong \ell^\infty$ by the usual mapping. $\endgroup$– Robert FurberCommented Apr 11, 2022 at 22:16
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$\begingroup$ This situation occurs in Solovay's model of ZF + DC with all subsets of $\mathbb{R}$ Lebesgue measurable, and also in Shelah's model with all subsets of $\mathbb{R}$ having the Baire property, but there are other ways of getting it. The upshot is that you need to use the Hahn-Banach theorem to prove that a closed subspace of a Banach space is reflexive. $\endgroup$– Robert FurberCommented Apr 11, 2022 at 22:19
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