Let $X$ be a separable Banach space such that given any other separable Banach space $Y$, there exists an isometry $j:Y \longrightarrow X$.
Now, let $\mathcal{F}(X)$ be the set of all closed sets of $X$ equipped with the Effros-Borel structure and $SB(X)$ the set of all closed subspaces of $X$.
I can then be proven that $SB(X)$ is standard Borel space, that is there exists a Polish topology on $SB(X)$ such that its Borel $\sigma$-algebra is exactly the set form by the intersection of $SB(X)$ with every set of the Effros-Borel structure on $\mathcal{F}(X)$.
Let $Z$ be any other separable infinite-dimensional Banach space and $\{z_n\}_{n=1}^\infty$ a sequence of linearly independent vectors in $Z$ such that $$Z=\overline{\operatorname{span}}\{z_n: n \in \mathbb{N}\}$$ I'm trying to prove that the set $$B=\left\{\left(\{f_n\}_{n=1}^\infty, F \right) \in X^\mathbb{N} \times SB(X): f_n \in F, \forall n \in \mathbb{N} \wedge \operatorname{span}\{f_n\} \simeq \operatorname{span}\{z_n\} \wedge \overline{\operatorname{span}}\{f_n\} = F \right\}$$ is a Borel set in $X^\mathbb{N} \times SB(X)$, but I'm completely sutcked so any help would be really appreciated.