In the absence of choice, is the countable union of translates of $\Bbb Q$ countable?

Question

It is a well-known theorem of $\sf ZFC$ that the countable union of countable sets is countable. That is, if $X_i$ is countable for $i\in\Bbb N$, then $\bigcup_iX_i$ is countable as well. However, this is not a theorem of $\sf ZF$. The issue is, essentially, that knowing that a set is countable is weaker than having a particular witness of that countability. To show that $\bigcup_iX_i$ is countable, one needs to choose an enumeration of each $X_i$.

What if each $X_i$ is a translate of $\Bbb Q$? That is, suppose $\{X_i\}_{i\in\Bbb N}$ is a countable collection of sets where for each $i$, there exists a $p\in\Bbb R$ such that $X_i=\Bbb Q+p$. In this case, does $\sf ZF$ prove that $\bigcup_iX_i$ is countable?

(Alternatively, we may specify, for each $i$ and for all $x,y\in X_i$, that $x-y$ is rational. This implies that each $X_i$ is a subset of a translate of the rationals, and so it is essentially the same question.)

I would write $p_i$ to make it clear $p$ depends on $i,$ and thus is necessarily a function. — Thomas Andrews, Apr 11 '25 at 19:54
@ThomasAndrews $p$ depends on $i$, but concluding that it is a function is exactly the content of the axiom of choice! In any case, the dependency is made clear from the order of the qualifiers. — Akiva Weinberger, Apr 11 '25 at 20:00
If you have an enumeration $y_j$ of $Y=\bigcup X_i,$ then we can define a function $i\mapsto p_i$ as $p_i=x_j$ where $j$ is the smallest natural number such that $x_j\in X_i.$ — Thomas Andrews, Apr 11 '25 at 20:00
So this means there is such an enumeration iff there exists a function $p_i.$ (That assumes the $X_i$ are known to be distinct, I think.) — Thomas Andrews, Apr 11 '25 at 20:07
Largely equivalent to https://math.stackexchange.com/questions/2456608/cardinality-of-the-union-of-a-countable-number-of-countable-well-ordered-sets — Eric Towers, Apr 11 '25 at 20:33
@EricTowers Not from what I can tell, since $(\Bbb Q,<)$ is not well-ordered. — Akiva Weinberger, Apr 11 '25 at 21:45
@AkivaWeinberger : Yes, even if it were well-ordered, so that each translate has a direct bijection with the integers, you still don't get the result you want. Without (well-)ordering, the situation is not stronger. — Eric Towers, Apr 11 '25 at 22:04

score 5 · Accepted Answer · answered Apr 12 '25 at 19:23

Here's a quick sketch for a negative answer.

Consider the forcing adding countably many Cohen reals. Unlike the usual presentation, add these by rational approximations of sorts.

Now, consider the symmetric system given by the automorphism group which shifts each Cohen real by a rational value, but does not permute the coordinates.

Take the filter of subgroups which fixes pointwise finitely many coordinates.

The resulting model has that the sequence of the Cohen reals is not in the model, but the sequence of their equivalence classes "up to a rational difference" is in the model. Each of these is a shift of the rationals by the corresponding Cohen real.

Consequently, the union of these equivalence classes is not a countable set, or else we could obtain the sequence of the Cohen reals.

Apologies, I'm on holiday so it took longer than necessary to post this answer. — Asaf Karagila, Apr 12 '25 at 19:24

In the absence of choice, is the countable union of translates of $\Bbb Q$ countable?

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