I want to show $c_0$ is not reflexive by showing the canonical embedding $c_0\rightarrow c_0^{**}$ is not surjective.
I saw a reference said: since the inclusion map $c_0\rightarrow\ell_\infty$ is not surjective, the canonical embedding $c_0\rightarrow {c_0}^{**}$ is not surjective.
I know $c_0$ is a closed linear subspace of $\ell_\infty$ and ${c_0}^{**}\cong{\ell_1}^*\cong\ell_\infty$. But how the canonical embedding is related to the inclusion map $c_0\rightarrow\ell_\infty$?
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 $\ell_\infty$ and hence less than ${c_0}^{**}$. 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 precise proof by using the reference idea? Thank you!
You have to use what the canonical embedding means. And, that you can identify $c_0^*$ with $\ell_1$.
If $a\in c_0$, you see it as an element of $c_0^{**}$ via $\hat a(b)=b(a)$, for each $b\in\ell_1$. That is, $$ \hat a(b)=\sum_na_nb_n. $$ This is precisely the duality $\ell_\infty=\ell_1^*$. In other words, if you are working with the characterizations $c_0^*=\ell_1$, and $\ell_1^*=\ell_\infty$, then the canonical embedding of $c_0$ in $c_0^{**}$ is precisely the embedding $c_0\subset\ell_\infty$.