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)?


  • 2
    $\begingroup$ 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


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.


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.

  • 1
    $\begingroup$ Turing reducible. $\endgroup$ Aug 12 at 12:37
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ @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

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 $r$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.