15
$\begingroup$

Let $P$ denote the following proposition:

There exists a set $S$ of subsets of $\mathbb{R}$ such that

  • $S$ is totally ordered by inclusion;
  • each member of $S$ has no accumulation points;
  • the union of $S$ is $\mathbb{R}$.

My question: Assuming the consistency of ZF, is $\,\text{ZFC}+P\,$ consistent?

NB: I'm pretty sure that $P$ is not provable in ZFC, as I think that $P$ implies the continuum hypothesis.

Improve this question
$\endgroup$
3
  • $\begingroup$ I think, this would not work even if you relaxed the point 2 to "each member of $S$ is countable" $\endgroup$
    – Anixx
    Commented Jun 27 at 21:25
  • $\begingroup$ Side remark: this appears to be related to the principle $\Phi$ introduced by Hathaway and I in arxiv.org/pdf/1601.05454 (see sections 2 and 3). $\endgroup$
    – François G. Dorais
    Commented Jun 27 at 23:35
  • $\begingroup$ @Anixx Your situation is equivalent to CH. If CH holds, $\mathbb{R}$ is the union of a tower of countable sets. The $\alpha$th set will be the reals below $\alpha$ in the enumeration of $\mathbb{R}$ in type $\omega_1$. And one can prove the converse as well. $\endgroup$
    – Joel David Hamkins
    Commented 2 days ago

2 Answers 2

Reset to default
22
$\begingroup$

It is a very nice question, but unfortunately, this is impossible.

Each member $s\in S$ must be countable, since uncountable sets have accumulation points. And since the hierarchy is accumulating as you go up, the cofinality of the order must be $\omega_1$ (and this is presumably how you deduced CH). It must be at least $\omega_1$, since otherwise we'd have $\mathbb{R}$ as a countable union of countable sets; and the cofinality cannot be larger than $\omega_1$, since otherwise we'd have eventually an uncountable set in the family.

So we may pass to a cofinal suborder, an $\omega_1$ increasing sequence of countable sets $s_\alpha$ with $\bigcup_{\alpha<\omega_1} s_\alpha=\mathbb{R}$, but no $s_\alpha$ has accumulation.

Since every real number shows up eventually, however, at some countable stage all the rationals will be present, but then every $s_\beta$ after that will have many accumulation points, contrary to assumption.

Improve this answer
$\endgroup$
2
  • 6
    $\begingroup$ Could argue directly, without mentioning $\omega_1$, like this: every rational appears in some $s\in S$, but they cannot appear together, so $S$ has a cofinal $\omega$ sequence. But since each $s$ is countable, the union of all of them will be countable, so it isn't $\mathbb{R}$. $\endgroup$
    – Joel David Hamkins
    Commented Jun 27 at 16:17
  • 1
    $\begingroup$ If you press my argument harder, you get the theorem: every lineared ordered family of sets of reals, each set in the family having no accumulation point, has a countable union. If it had an uncountable union, then that union would have an accumulation point, and so some countable set of points in the union would have an accumulation point, but they would have to all appear by some stage, or else the cofinality would be countable and the union would be countable. $\endgroup$
    – Joel David Hamkins
    Commented Jun 27 at 21:02
12
$\begingroup$

Here's a ZF proof that if $S$ is a chain of sets with $\bigcup S = \mathbb{R},$ then there is $X \in S$ which contains a countable set dense in some nonempty open set.

If there is $X \in S$ such that $\mathbb{Q} \subset X,$ we are done. Assume this is not the case. Let $\langle q_n \rangle$ be an enumeration of $\mathbb{Q}.$ Let $$X_n = \bigcap \{X \in S: \forall i<n (q_i \in X)\}.$$ Let $Y_n = X_{n+1} \setminus X_n.$ Then $\langle Y_n \rangle$ is a countable partition of $\mathbb{R}.$ I show in the first part of my answer here that some $Y_n$ contains a countable dense subset of an interval. Then any $X \in S$ containing $\{q_0, \ldots, q_n\}$ is as desired. $\square$

Generalizing along a different axis, if $S$ is a chain of subsets of $\mathbb{R}$ each discrete under the subspace topology, then $\bigcup S$ is countable. Let $\langle U_n \rangle$ enumerate a topological basis. Then we can construct a surjective partial map $f: \omega \rightharpoonup \bigcup S$ by $f(n) = x$ if there is $X \in S$ such that $U_n \cap X = \{x\}.$

Both of the above readily generalize to an arbitrary Polish space.

Improve this answer
$\endgroup$

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