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!



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


