Bessaga-Pelczynski Selection Principle states that if $(x\_{n})\_{n}$ is a basis for a Banach space $X$, then every normalized weakly null sequence $(y\_{n})\_{n}$ in $X$ admits a subsequence that is equivalent to a block basic sequence with respect to $(x\_{n})\_{n}$. I am wondering whether the sequence $(y\_{n})\_{n}$, for every $\epsilon>0$, admits a subsequence sequence $(y\_{k\_{n}})\_{n}$ that is $(1+\epsilon)$-equivalent to a block basic sequence with respect to $(x\_{n})\_{n}$. Thank you!



Bessaga-Pelczynski Selection Principle
If $(x\_n)\_{n=1}^\infty$ is normalized then yes, that is true. It is even true if we replace the normalization criterion with $\lim\|y\_n\|=1$. The proof is in Albiac-Kalton, although it is not formally stated there. For a reference, see Theorem 1.1 [here](http://www.eweb.unex.es/eweb/extracta/Vol-29-1-2/29J1_2Wallis.pdf). (Note that we also require $(x\_n)\_{n=1}^\infty$ to satisfy $\lim\|x\_n\|=1$, even though that is not stated in the reference. But I assume you are using a normalized $(x\_n)\_{n=1}^\infty$, so this is not an issue.)


