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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.

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An obvious point that might as well be stated: $\Gamma$ must have cardinality at most $2^{\aleph_0}$, simply for an injective map to exist. – Jonas Meyer Aug 26 '11 at 18:50
Perhaps also worth mentioning is that it is possible to have an injective bounded linear operator from a nonseparable Banach space to $c_0$. E.g., define $T:\ell^\infty\to c_0$ by $T(x_0,x_1,x_2,\ldots)=(x_0,\frac{1}{2}x_1,\frac{1}{3}x_2,\ldots)$. – Jonas Meyer Aug 26 '11 at 19:01
Thanks, however $c_0(\omega_1)$ does not embed into $\ell^\infty$ (unlike $\ell^1(2^{\aleph_0})$). This is by Pełczyński's theorem: $\ell^{\infty}(\omega_1)$ would have to embed into $\ell^\infty$ as well which is not the case. – F.R. Rogers Aug 26 '11 at 19:13

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 $c_0(\omega_1)$).

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One day, if I have time, I might distill Olagunju's argument and explicate it here, as it's not that long or complicated. – Philip Brooker Sep 9 '11 at 8:33

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.

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$T^{-1}$ is not continuous. You also seem to assum that $T(c_0(\Gamma))$ is closed. Am I misunderstanding? – Jonas Meyer Aug 26 '11 at 18:37
yep, you're right I was trying to apply Banach inverse theorem, but for this the assumption that $T(c_0(\Gamma))$ is closed is required. – Franz Aug 26 '11 at 18:45

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