In the absence of AC, can there be an uncountable dense set $S\subset\mathbb R$ such that $S\cap(-\infty,a)$ is countable for each real number $a$?

(Of course, since $S$ is a countable union of countable sets, the countable axiom of choice must fail in order for $S$ to be uncountable.)


1 Answer 1


Sure — this will happen in any model where there is a countable set $X=\{A_i: i\in\mathbb{N}\}$ of countable sets of reals, whose union is not countable. Fix bijections $f_i$ from $\mathbb{R}$ to $(i, i+1)$ for each $i\in\mathbb{N}$, and let $$\hat{S}=\bigcup_{i\in\mathbb N} f_i(A_i).$$ Now to make it dense, just let $S=\hat{S}\cup\mathbb{Q}$. (Lest it seem like I'm sweeping something under the rug here, note that such $f_i$s can be explicitly described: $f_i(x)=i+{{\pi\over 2}+\arctan(x)\over\pi}$.)

Meanwhile, we can get models where a countable union of countable sets of reals isn't countable in a number of ways: the most fun (YMMV) is to look at a model where $\mathbb{R}$ itself is a countable union of countable sets!


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