Is it consistent with ZF that $V \to V^{\ast \ast}$ is always an isomorphism?

Let $k$ be a field and $V$ a $k$-vector space. Then there is a map $V \to V^{\ast \ast}$, where $V^{\ast}$ is the dual vector space. If we are in ZFC and $\dim V$ is infinite, then this map is not surjective. As we learned in this question, there are models of $ZF$ where $V \to V^{\ast \ast}$ is an isomorphism when $V$ has a countable basis. I think the same argument shows that it is consistent with ZF that this is an isomorphism whenever $V$ has a basis.

Is it consistent with $ZF$ that $V \to V^{\ast \ast}$ is an isomorphism for all vector spaces $V$?

I ask because I'm teaching a rigorous undergrad analysis class. My students keep asking me whether they have to believe that $V \to V^{\ast \ast}$ can fail to be an isomorphism. Of course, I'm trying to change their intuition to point out why most mathematicians find the failure of isomorphism plausible and point out that there are more subtle ways to salvage the claim, such as Hilbert spaces, but I'd also love to be able to give them a choice free proof that there is some vector space where this issue comes up.

• This is a question I asked myself some years ago, which is why I had an answer waiting. A related question that I also asked myself, although I’m afraid I don’t remember whether I came to any conclusion, is whether it’s possible that $V\to V^{**}$ is always surjective. Of course, “always injective” is possible, as that’s what happens if the axiom of choice is true. – Jeremy Rickard Jan 11 '18 at 18:00

No, it’s not consistent.

Let $V=k^{(\omega)}$ be the vector space of finite sequences of elements of $k$. Then $V^*$ can be identified with the vector space $k^\omega$ of all sequences, and elements of the image of the natural map $V\to V^{**}$, considered as maps $k^\omega\to k$, are determined by their restriction to $k^{(\omega)}$.

So if $V\to V^{**}$ is an isomorphism, then, taking $W=k^\omega/k^{(\omega)}$, there are no nonzero linear maps $W\to k$, and hence $W^{**}=0$. But $W$ is nonzero.

So the map to the double dual must fail to be an isomorphism either for $V$ or for $W$.

• Sorry if I’m missing something obvious, but how do you prove without the axiom of choice that any nontrivial linear space carries a nonzero linear map to $k$? – Emil Jeřábek supports Monica Jan 11 '18 at 15:32
• @EmilJeřábek He doesn't. The argument breaks down into two cases. Case 1: $W$ has a nonzero map to $k$. Then the composition $V \to W \to k$ is an element of $V{\ast \ast}$ not in $V$. Case 2: $W^{\ast}=0$. Then $W^{\ast \ast}=0$, so $W \to W^{\ast \ast}$ has a kernel. – David E Speyer Jan 11 '18 at 16:04
• Ah! You are right, I didn't read the proof carefully enough. – Emil Jeřábek supports Monica Jan 11 '18 at 16:18
• @AaronBergman Hmmm... If I’m forced to choose, then can I cheat and pick $V\oplus W$? :) – Jeremy Rickard Jan 12 '18 at 8:25
• @DavidLampert No. Turns out I can make jokes without even trying! – Jeremy Rickard Jan 12 '18 at 18:04

Not an answer to your question but a variant which is perhaps more à propos given that you said you are teaching a rigorous undergrad analysis class.

Let $V=\oplus_{\mathbb{N}}\mathbb{R}$ be the space with countable basis which motivated your question. Another way to salvage the isomorphism $V\simeq V^{\ast\ast}$ (without giving up any cherished axiom) is to think of $V$ as equipped with the finest locally convex topology and consider duals as topological duals always given the strong topology. It is easy to see that $V^{\ast}\simeq \prod_{\mathbb{N}}\mathbb{R}$ with the product topology. Although perhaps counterintuitive, when one takes the proper (i.e., strong topological) dual of $\prod_{\mathbb{N}}\mathbb{R}$, then one gets back to $\oplus_{\mathbb{N}}\mathbb{R}$.