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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.2011-08-26
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    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 $T(x_0,x_1,x_2,\ldots)=(x_0,\frac{1}{2}x_1,\frac{1}{3}x_2,\ldots)$.2011-08-26
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    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.2011-08-26

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