I want to show $c_0$ is not reflexive by showing the canonical embedding $c_0\rightarrow c_0^{**}$ is not surjective. I know there are some ways to get it. But I am thinking can I use cardinality to show the result.

And my attempt is: since ${c_0}^{**}\cong{\ell_1}^*\cong\ell_\infty$, there exists a bijection $T:\ell_\infty\rightarrow {c_0}^{**}$. Then they have the same cardinality. Since $c_0\rightarrow\ell_\infty$ is not surjective, the cardinality of $c_0$ is less than the cardinality of $\ell_\infty$ and hence less than the cardinality of ${c_0}^{**}$. Thus the canonical embedding $c_0\rightarrow {c_0}^{**}$ is not surjective. But I am not sure it is true and I feel that I missed something.

Can anyone give me a clue? Thank you!

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    $\begingroup$ Both, $c_0$ and $c_0^{\ast\ast}$ have cardinality $\mathfrak{c} = 2^{\aleph_0}$. Cardinality doesn't work. $\endgroup$ – Daniel Fischer Apr 12 '18 at 17:57
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    $\begingroup$ To say $A$ has cardinality less than $B$ you need to show that there is no surjective map from $A$ to $B$: one example of a non-surjective map is not enough. Say $A$ is the integers and $B$ is the even integers; define $f:A\to B$ by $f(n)=4n$. Then $f$ is not surjective, but $A$ and $B$ have the same cardinality. $\endgroup$ – David C. Ullrich Apr 12 '18 at 18:01

The smallest cardinality of a dense subset (known as the density character of a Banach space) can be used here: $c_0$ has a dense countable set (is separable) while $\ell_\infty$ does not.

The cardinality itself does not distinguish sequence spaces from one another; they are all in bijection with one another.


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