Non-reflexive module isomorphic to its double dual Could you give me an example of a non-reflexive module isomorphic to its double dual?


I found an example here but I cannot understand it, do you have any simpler examples?

By this question we should only use not finitely generated modules.
What about $l^1(\mathbb{R})$ as an $\mathbb{R}$ vector space?
The natural map $J_1\colon l^\infty \to (l^1)^\vee$, which reads a bounded sequence as the multiplication by said sequence, is an isomorphism, while the dual map $J_2\colon l^1 \to (l^\infty)^\vee$ is not. We then get, composing the natural bidual map with the traspose of $J_1$, that
$$l^1 \to (l^1)^{\vee\vee} \to (l^\infty)^{\vee}$$
is, again, $J_2$! Hence $l^1$ cannot be reflexive. Still, we can find bases of the same cardinality for the two spaces.
What about $l^1(\mathbb{R})$ as an $\mathbb{R}$ vector space?
The natural map $J_1\colon l^\infty \to (l^1)^\vee$, which reads a bounded sequence as the multiplication by said sequence, is an isomorphism, while the dual map $J_2\colon l^1 \to (l^\infty)^\vee$ is not. We then get, composing the natural bidual map with the traspose of $J_1$, that
$$l^1 \to (l^1)^{\vee\vee} \to (l^\infty)^{\vee}$$
is, again, $J_2$! Hence $l^1$ cannot be reflexive. Still, we can find bases of the same cardinality for the two spaces.