Using the (countable) axiom of choice, one can show that a countable union of countable sets is again countable. I was wondering whether we need the axiom of choice in the following specific case of a countable union (which I encountered reading about the Banach-Tarski paradox).

Let $(X_i)_{i \in I}$ be a countable collection of countable subsets of $\mathbb{R}$. Can it be proven in ZF that the union $\bigcup_i X_i$ is countable?

I have thought about some `explicit' injections of the $X_i$ into the natural numbers, using it is a subset of the reals. For example, in the specific case that all $X_i$ are discrete subsets of $\mathbb{R}$, we can construct a collection of injections $(X_i \to \mathbb{Z})_i$ (and then it follows that the union of the $X_i$ is countable). I think the general case could be done if there exists a choice function on the non-empty countable subsets of $\mathbb{R}$. I also don't know whether the existence of such a function can be shown in ZF.


1 Answer 1


No. It is consistent with $\sf ZF$ that $\Bbb R$ is a countable union of countable sets.

This is a result due to Feferman and Levy from 1963, you can find the details in Jech's "The Axiom of Choice" in Chapter 10.


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