I am currently studying a functional analysis course as part III of the Cambridge Tripos. In the course we were quoted the result:
Let $SB$ be the class of separable Banach spaces over $\mathbb{R}$.
Let $Z$ be a $\mathbb{R}$ normed vector space in which $\forall X \in SB \; \exists\; Y \subset Z $ and $T:X \to Y $ isomorphism ,then $\exists c>0$ such that $\forall X \in SB \; \exists\; Y \subset Z $ and $T:X \to Y $ isomorphism with $||T||,||T^{-1}|| \leq c$.
I'm trying to prove this example but I'm not quite sure where to begin. The only similar result that we have proved in lectures is that every separable Banach spaces can be isometrically embedded into $l^{\infty}$. Here $$ l^{\infty}=\{(x_{n})_{n\in \mathbb{N}} \subset \mathbb{R}:sup_{n \in \mathbb{N}}|x_{n}|<+ \infty\} $$ is the space of bounded real sequences.
Any help would be greatly appreciated.
Suppose there is no such $c>0$. Then, for each $n\in\mathbb{N}$ there exists separable Banach spaces $X_n$ such that for all isomorphisms $T_n:X_n\to T_n(X_n)\subseteq Z$, we have either $\|T_n\|\geq n$ or $\|T_n^{-1}\|\geq n$. Thus, either there exists a subsequence $(X_{n_k})$ for which $\|T_{n_k}\|\geq n_k$, or a subsequence $(X_{n_k})$ for which $\|T_{n_k}^{-1}\|\geq n_k$.
WLOG, assume $\|T_{n_k}\|\geq n_k$ for a subsequence (the proof of the other case is similar). Let $$E = \left(\bigoplus_{k\in\mathbb{N}} X_{n_k}\right)_{\ell^1} = \{(u_k)_{k\in\mathbb{N}} : \sum_{k\in\mathbb{N}} \|u_k\|_{X_{n_k}}<\infty \}$$ with the norm $$\|(u_k)_{k\in\mathbb{N}}\|_E = \sum_{k\in\mathbb{N}} \|u_k\|_{X_{n_k}} .$$ Then, $E$ is a separable Banach space. By hypothesis, there exists an isomorphism $T:E\to T(X)\subseteq Z$.
For each $k\in\mathbb{N}$, $E$ contains a natural closed subspace, which is isometrically isomorphic to $X_{n_k}$. WLOG, call this subspace also $X_{n_k}$. The restriction of $T$ to $X_{n_k}$ provides an isomorphism, so we must have $$\|T\| = \sup_{u\in E} \frac{\|Tu\|}{\|u\|} \geq \sup_{u\in X_{n_k}} \frac{\|Tu\|}{\|u\|} \geq n_k.$$ Contradiction.