# Consistency of a strange (choice-wise) set of reals

Consider a set $$X\subseteq \mathbb{R}$$ such that

1. $$X$$ is not separable wrt its subspace topology
2. For all $$r\in\mathbb{R}$$ there exists a sequence $$(x_n)_{n\in\omega} \subset X$$ converging to $$r$$

In a model containing such a set $$\text{AC}_\omega(X)$$ (choice for countable families of non-empty subsets of $$X$$) would of course fail, but not that critically.

For example, the unique way I've seen to prove the consistency of the existence of a non-separable set of reals is the one that shows the consistency of an infinite, Dedekind-finite set of reals, but our set, though being non-separable, is well-behaved enough to witness density in a sequencial manner.

My questions are:

• Is its existence consistent relative to $$\text{ZF}$$? Has it been proved somehwere?
• In case the answer above is "no", does this remind you similar results (besides the most known ones that can be found in Jech' Axiom of Choice)?

Thanks!

• I don't think I've seen that result. As an attempt, I'd try adding Cohen reals indexed by $\Bbb Q$, take order preserving automorphisms, and generate the filter by fixing pointwise a sequence, or a bounded sequence. The resulting set of reals is going to be dense in some interval at least, so we can stretch it to get a set dense in the reals. So it is enough to show that in its dense area it satisfies (2), which should be doable, I think. Finally, to show it's not separable, I'm not sure if the usual trick is going to work, but it seems like there might be a way to solve that. Aug 12 at 10:06
• @AsafKaragila The support of the Cohen product is finite or countable (full since we are indexing by $\mathbb{Q}$)? Aug 12 at 10:18
• Finite. It's just adding Cohen reals. Indexing by $\Bbb Q$ makes it conceptually easier to understand the automorphisms, that's all. Aug 12 at 10:54

The existence of such a set follows from $$\mathbb{R}$$ is a countable union of countable sets.$$"$$ Let $$\mathbb{R} = \bigcup_{n<\omega} S_n,$$ each $$S_n$$ countable. Let $$T_n = \{x \in \mathbb{R}: \exists m \le n \exists y \in S_m (x \le_T y)\}$$ and $$X_n = (2^{-n-1}, 2^{-n}) \setminus T_n.$$

We will show $$X = \bigcup_{n<\omega} X_n$$ is a subset of $$(0,1)$$ with the desired properties. Condition (2) follows from $$X$$ having cocountable intersection with each $$(2^{-n}, 1).$$ Suppose condition (1) fails. Let $$r \in S_n$$ encode a dense sequence $$\langle r_i: i<\omega \rangle \subset X.$$ Then $$r_i \in T_n$$ for all $$i,$$ so $$2^{-n} contradiction.

Edit:

It turns out the nonexistence of such a set is equivalent to $$\text{AC}_{\omega}(\mathbb{R}).$$ First, $$\text{AC}_{\omega}(\mathbb{R})$$ implies every set of reals is separable. For the other direction, suppose $$\langle S_n: n<\omega \rangle$$ is a sequence of nonempty sets of reals without a choice function. Let $$T_n = \{x \in \mathbb{R}: \forall m \le n \exists y \in S_m (y \le_T x)\}$$ and $$X_n = (2^{-n-1}, 2^{-n}) \cap T_n.$$

We will show $$X = \bigcup_{n<\omega} X_n$$ is a subset of $$(0,1)$$ with the desired properties. Condition (2) follows from the fact that each $$T_n$$ is nonempty and closed under addition by rational numbers. Suppose condition (1) fails. Let $$r$$ encode a dense sequence $$\langle r_i: i<\omega \rangle \subset X.$$ Then $$\{x \in \mathbb{R}: x \le_T r\}$$ is a countable set which meets each $$S_n,$$ which contradicts the fact that $$\langle S_n \rangle$$ has no choice function.

• Turing reducible. Aug 12 at 12:37
• So what you're really just looking for is "countable unions of countable sets of reals could be uncountable", which I think would turn out equivalent. This is also equivalent to the existence of an $\aleph_1$-amorphous set of reals, if my memory serves me right. Eilon Bilinsky had some work on that subject published in Proc. AMS a few years ago. Aug 12 at 12:56
• @AsafKaragila It's not clear to me if that's enough. That $\langle r_i \rangle$ is encoded by a real in some $S_n$ depends on their union being all of $\mathbb{R}.$ Aug 12 at 12:59
• Thanks, just one question: why is the fact that $X$ has cocountable intersections with $(2^{-n},1)$ sufficient to find the converging sequences? Aug 12 at 13:00
• @Lorenzo From an enumeration of $(2^{-n}, 1) \setminus X$ one can use diagonalization to canonically choose a real from each $X \cap I$ ($I$ a subinterval). Aug 12 at 13:03

Here is another way to show the consistency of such a set by a direct symmetric extension approach:
Let $$\mathbb{P}$$ be the forcing that add Cohen reals (by reals I mean elements of $$\omega^\omega$$) indexed by $$\omega\times\omega$$ and let $$\mathcal{G}$$ be the group of all permutations of $$\omega\times\omega$$ such that for every $$n$$, $$\pi (n,i) = (n,j)$$ for some $$j$$.
Let $$\mathcal{F}$$ be the filter on $$\mathcal{G}$$ generated by $$\{H_n \mid n \in \omega\}$$ where $$H_n$$ consists of all $$\pi$$ such that $$\pi (k,i) = (k,i)$$ for all $$k\le n$$, all $$i\in\omega$$.

Let $$x_{k,i}$$ be the Cohen reals added in the symmetric extension and $$A_k = \{x_{k,i} \mid i \in \omega\}$$, then the function $$k \mapsto A_k$$ is in the symmetric extension and so is $$A = \bigcup_{k} k^\smallfrown A_k$$, where $$x \in k^\smallfrown A_k$$ if $$x(0) = k$$ and there exists $$y \in A_k$$ such that $$x(n+1) = y(n)$$ for all $$n$$.
Now $$A$$ is not separable, because otherwise there would be a choice function for $$k\mapsto A_k$$ (which doesn't exists in our symmetric model by construction) but each $$A_k$$ is separable and dense, therefore, given any $$r\in\omega^\omega$$, if $$r(0) = k$$ then I can find a sequence in $$k^\smallfrown A_k$$ converging to $$r$$.