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!
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.