Does there exist a model of ZF in which every linear functional on $\prod_{n \in \Bbb N} \Bbb F_2$ is continuous?

Question

I should preface this question by saying that I know almost no serious set theory.

Inspired by an answer of mine I got curious about a set-theoretic problem: Equip the $\Bbb F_2$ vector space $\prod_{n \in \Bbb N} \Bbb F_2$ with the product topology. Is there a model of ZF where every linear map $\prod_{n \in \Bbb N} \Bbb F_2 \to \Bbb F_2$ is continuous?

The motivation for this question comes from the linked answer: there I used an a choice-dependent argument to produce a discontinuous functional on this vector space to give an example of a finite index group in a profinite group that is not open. The problem of finding such examples is relevant (among other things) because a celebrated theorem (using the classification of finite simple groups!) says that there are no such examples for topologically finitely generated profinite groups. Asking for every finite index subgroup of a profinite group to be open is equivalent to asking if a group is its own profinite completion (as an abstract group), which is a natural question to ask about profinite groups.

What I have tried (despite knowing no set theory): I have a feeling this might be related to ultrafilters, i.e. that in a model of ZF with no non-principal ultrafilters there might be no discontinuous functional. The reason I believe this is that if we consider $\prod_{n \in \Bbb N} \Bbb F_2$ as a ring with pointwise multiplication, then multiplicative functionals $\prod_{n \in \Bbb N} \Bbb F_2 \to \Bbb F_2$ correspond to ultrafilters and continuous ones correspond (I think) to principal ultrafilters. But of course this isn't an answer, as not every functional is multiplicative.

Answer 1

I have no idea if it is related to ultrafilters (although it sounds plausible that if the ultrafilter lemma holds, then there is a discontinuous linear functional from $\mathbb{F}_2^\omega$ to $\mathbb{F}_2$) but let me prove that Solovay model thinks every linear functional $\mathbb{F}_2^\omega \to \mathbb{F}_2$ is continuous.

We can equip $\mathbb{F}_2^\omega$ a natural topology making the vector space homeomorphic to the Cantor space. Its topology is generated by $B_s$ for $s\in \mathbb{F}_2^{<\omega}$, where $$B_s = \{v\in \mathbb{F}_2^\omega\mid s\subseteq v\}.$$

We can impose a complete measure (whose construction is similar to that of Lebesgue measure over $[0,1]$) $\mu$ satisfying $\mu(B_s)=2^{-|s|}$. It is known that Solovay model satisfies $\mathsf{ZF+DC}$ and every subset of $\mathbb{F}_2^\omega$ is $\mu$-measurable. $\mu$ also enjoys various properties what Lebesgue measure has.

We claim (over $\mathsf{ZF+DC}$) that if a linear functional $F\colon \mathbb{F}_2^\omega \to \mathbb{F}_2$ is $\mu$-measurable, then it is continuous at the zero vector, which implies the continuity of $F$. Its proof is more or less similar to the real number case, but let me rephrase in our setting.

By Lusin's theorem, we have a compact set $K\subseteq \mathbb{F}_2^\omega$ such that $\mu(K)>2/3$ and $F$ is uniformly continuous over $K$. Thus we can pick $n\in\mathbb{N}$ such that if $x\restriction n=y\restriction n$, then $F(x)=F(y)$ for $x,y\in K$.

Now let $z\in\mathbb{F}^\omega_2$ be any vector such that the first $n$ components are zero. Then we can see that $K \cap (K+z)\neq\varnothing$, otherwise we have a contradiction by the following inequality: $$1 = \mu(\mathbb{F}^\omega_2)\ge \mu(K)+\mu(K+z)\ge 4/3.$$ Hence we can find $x_0\in K\cap (K+z)$, so $x_0+z\in K$. It shows $F(x_0)=F(x_0+z)$, so $F(z)=0$.

In sum, we found $n\in\mathbb{N}$ such that for every $z\in\mathbb{F}^\omega_2$, if $z\restriction n = \mathbf{0}\restriction n$, then $F(z)=0$, which is the definition of the continuity of $F$.