Injection from non-separable to separable subspaces Let $\Gamma$ be an uncountable set (possibly of cardinality $\aleph_1$). Is there an injective bounded linear operator $T\colon c_0(\Gamma)\to X$, where
a) $X$ is some separable Banach space
b) $X=c_0$?
Thank you in advance.
EDIT: This might be useful as well: Johnson and Zippin proved that each quotient of $c_0$ is in fact its subspace. Is there a similiar result for general $c_0(\Gamma)$ spaces?
EDIT 2: Another hint: If $T_1\colon Y\to c_0(\Gamma)$ is injective, then there exists an injective operator $S\colon Y\to c_0(\Gamma)$ with dense range.
 A: The answer is no in both cases. To see why, I suggest first observing that every separable Banach space $X$ admits a one-to-one continuous linear mapping into Hilbert space; consider, e.g., an isomorphic embedding of $X$ into $\ell_\infty$ composed with a one-to-one diagonal operator from $\ell_\infty$ into $\ell_2$. It thus suffices to show that $c_0(\omega_1)$ does not admit a one-to-one continuous linear map into any Hilbert space; for this, there may be many references, but the one that comes to mind for me is Olagunju's paper A Banach space that cannot be made into a BIP space, Proc. Camb. Phil. Soc. 63 (1967), 949-950 (Olagunju actually shows that $\ell_\infty([0,1])$ does not admit a one-to-one continuous linear operator into any Hilbert space, but you will see that the same argument works for $c_0(\omega_1)$).
A: Assume that there exists such an injective bounded linear operator $T \colon c_0(\Gamma) \rightarrow X$ that a) is satisfied. Then $T \colon c_0(\Gamma) \rightarrow T(c_0(\Gamma))$ is bijective. Since $T(c_0(\Gamma))$ is a subspace of $X$ it is separable. Let  $Q$ be a countable set such that $T(c_0(\Gamma))= \text{cl}Q$. For the inverse $T^{-1} \colon T(c_0(\Gamma)) \rightarrow c_0(\Gamma)$ which is a continuous bijection we have that $T^{-1}(Q)$ is countable and it is  dense set in $c_0(\Gamma)$ which is a contradiction since $c_0(\Gamma)$ is not separable. 
Thus a) and b) cannot be satisfied. 
