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

Question

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!

Answer 1

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 $2^{-n}<r_i,$ 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.

Answer 2

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 $r$.