Let $E$ be a Banach space with basis. Is it possible to find examples such that $S \subset E$ be a complemented subspace without basis?
A complemented subspace without basis of a space with a basis
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functional-analysis
banach-spaces
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0[This](https://www.jstor.org/stable/2048451?seq=1#page_scan_tab_contents) looks relevant. – 2017-02-19
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0The paper of Szarek, mentioned in the above link (and giving an affirmative answer to your question) is: Szarek, S. J. A Banach space without a basis which has the bounded approximation property. Acta Math. 159 (1987), no. 1-2, 81-98. – 2017-02-19
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Yes. The paper: Szarek, S. J. A Banach space without a basis which has the bounded approximation property. Acta Math. 159 (1987), no. 1-2, 81-98, provides an example heralded by the title.
A separable Banach space $X$ has the bounded approximation property if and only if $X$ is isomorphic to a complemented subspace of a space with a basis. This result can be found in Classical Banach Spaces, Vol.1, Lindenstrauss and Tzafriri (Theorem 1.e.13).