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Is there a total ordering on $\mathcal P(\Bbb R)$, the set of all subsets of $\Bbb R$, such that the set of countable subsets is dense in it?

(Given a total ordering $(X,>)$, a set $A\subseteq X$ is dense in $X$, if, for any two distinct $x,y\in X$, there is an $a\in A$ such that $x<a<y$.)

I think that I could make something work if I use the well-ordering theorem on $\Bbb R$, but I have no idea how to do it without the axiom of choice. I don't even know if any total ordering on $\mathcal P(\Bbb R)$ exists without choice.

Akiva Weinberger
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Suppose the continuum hypothesis is true, and let $\prec$ be a well-ordering of $\mathbb{R}$ with order-type $\omega_1$. Consider the induced lexicographic order on $\mathcal{P}(\mathbb{R})$: that is, $A<B$ iff there is some $x\in \mathbb{R}$ such that $x\in B$, $x\not\in A$, and for all $y\prec x$, $y\in A$ iff $y\in B$. Then the countable subsets of $\mathbb{R}$ are dense. Indeed, if $A<B$ and $x\in\mathbb{R}$ is as above, let $C=B\cap \{y:y\preceq x\}$. Then $C$ is countable and $A<C\leq B$.

More generally, for any infinite cardinal $\kappa$, this gives a total ordering on $\mathcal{P}(\kappa)$ with a dense subset of size $2^{<\kappa}$ (namely, the lexicographic order and the subsets of cardinality $<\kappa$). In particular, if $\mathfrak{c}=2^{<\mathfrak{c}}$, this gives an ordering on $\mathcal{P}(\mathbb{R})$ with a dense subset of size $\mathfrak{c}$, which can then be identified with the countable subsets of $\mathbb{R}$.

Eric Wofsey
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  • You don't need to assume the continuum hypothesis. You just need the fact that $|\Bbb R|$ is equal to some ordinal, and this follows from the well-ordering theorem. Right? (Or are you saying that we don't need choice as long as the continuum hypothesis is true?) – Akiva Weinberger Sep 04 '15 at 02:29
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    @columbus8myhw: No, you just get the subsets of size $<2^\omega$ dense in the order, as in Eric’s second paragraph. – Brian M. Scott Sep 04 '15 at 05:13
  • Also, what about the finite subsets of $\Bbb R$? Are they also dense? – Akiva Weinberger Sep 04 '15 at 14:05