In a comment to a recent question, Jeremy Rickard asked whether it is consistent with ZF that the map $V \to V^{**}$ from a vector space to its double dual is always surjective. We know that "always injective" is consistent (since that's what happens in ZFC) and Jeremy Rickard's argument shows that "always an isomorphism" is *not* consistent. But what about "always surjective"?

Gro-Tsen points out that Harry West has already showed this is impossible elsewhere on MO. I am missing one step in West's answer. I thought I'd write up the issue here, and someone can explain to me what I am missing.

First of all, for any field $F$ and any set $X$, we can form the free vector space with basis $X$, call it $FX$, and the vector space of functions from $X$ to $F$, call it $F^X$. It is easy to see that $(FX)^{\ast} \cong F^X$ so, if $V \to V^{\ast \ast}$ is always surjective, then the obvious injection $FX \to (F^X)^{\ast}$ must always be an isomorphism. So we may and do assume:

**Key Consequence** For every set $X$, the obvious injection $FX \to (F^X)^{\ast}$ is an isomorphism.

Now, suppose that $\alpha$ is an ordinal with cofinality $>\omega$.

**Lemma 1** Let $X \subset \alpha$ have the property that $X \cap \beta$ is finite for every $\beta < \alpha$. Then $X$ is finite.

**Proof** If not, then $X$ is an infinite well-ordered set (by restricting the order from $\alpha$) so it contains a copy of $\omega$. By the hypothesis on $\alpha$, there is some $\beta_0 < \alpha$ containing this copy of $\omega$. But then $X \cap \beta_0$ is infinite. $\square$

Let $V$ be the subspace of $F^{\alpha}$ consisting of functions which are supported on $F^{\beta}$ for some $\beta< \alpha$.

**Lemma 2** $V^{\ast} = F \alpha$.

**Proof:** Let $\phi \in V^{\ast}$. We can restrict $\phi$ to $F^{\beta}$ for each $\beta < \alpha$ and, by the definition of $V$, the functional $\phi$ is determined by the list of these restrictions. On each $F^{\beta}$, by the Key Consequence, $\phi|_{F_{\beta}}$ coincides with some unique vector from $F \beta$. Let the support of that vector be $X_{\beta}$ and let $X = \bigcup_{\beta} X_{\beta}$. Then $X \cap \beta= X_{\beta}$ for each $\beta < \alpha$ so, by Lemma 1, $X$ is finite. The functional $\phi$ is then induced by a functional in $FX \subset F \alpha$. $\square$.

But then $V^{\ast \ast} = (F \alpha)^{\ast} = F^{\alpha}$, whereas $V$ is a proper subspace of $F^{\alpha}$. **QED**

My only issue is, in ZF, are we sure that there are ordinals of cofinality $>\omega$? In ZFC, one simply takes the first uncountable ordinal, $\omega_1$. If we had a cofinal sequence $0=x_0$, $x_1$, $x_2$, \dots, in $\omega_1$, then $\omega_1$ would be the union of the countable intervals $[x_i, x_{i+1})$, and would hence be countable.

But in ZF, a countable union of countable sets doesn't have to be countable. I tried some tricks to get around this and failed; please let me know what I missed.

6more comments