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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!

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    $\begingroup$ 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. $\endgroup$
    – Asaf Karagila
    Aug 12 at 10:06
  • $\begingroup$ @AsafKaragila The support of the Cohen product is finite or countable (full since we are indexing by $\mathbb{Q}$)? $\endgroup$
    – Lorenzo
    Aug 12 at 10:18
  • $\begingroup$ Finite. It's just adding Cohen reals. Indexing by $\Bbb Q$ makes it conceptually easier to understand the automorphisms, that's all. $\endgroup$
    – Asaf Karagila
    Aug 12 at 10:54

2 Answers 2

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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.

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    $\begingroup$ Turing reducible. $\endgroup$ Aug 12 at 12:37
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    $\begingroup$ 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. $\endgroup$
    – Asaf Karagila
    Aug 12 at 12:56
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    $\begingroup$ @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 $\mathbb{R}.$ $\endgroup$ Aug 12 at 12:59
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    $\begingroup$ Thanks, just one question: why is the fact that $X$ has cocountable intersections with $(2^{-n},1)$ sufficient to find the converging sequences? $\endgroup$
    – Lorenzo
    Aug 12 at 13:00
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    $\begingroup$ @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). $\endgroup$ Aug 12 at 13:03
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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 $r$.

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