Given a countable union of countable sets in $\mathbb{R}$, must at least one of them be dense in $\mathbb{R}$?

Question

Given a countable union of sets in $\mathbb{R}$ such that $\cup A_i = \mathbb{R}$, must at least one of them be dense in $\mathbb{R}$? And, if the answer is yes, can anyone tell how I can prove it?

Answer 1

No. For each prime $p_j$ consider the sets $S_j=\{p_j^k:k\in\mathbb{Z}^+\}$. Each of these sets is countable and disjoint. Their union does not even cover the positive integers.

---

Modified Question

We cannot deduce that one of the $A_i$ is dense, but we can get that one must be dense somewhere; that is, $\overline{A_i}$ contains an interval.

Suppose that none of the $A_i$ are anywhere dense. Then $\overline{A_i}$ contains no intervals, that is $\overline{A_i}^\complement$ is open and dense. The Baire Category Theorem says that $$ \bigcap_i\overline{A_i}^\complement=\left(\bigcup_i\overline{A_i}\right)^\complement=\mathbb{R}^\complement=\{\} $$ is dense (contradiction). Thus, one of the $A_i$ must be somewhere dense.

Note that given an interval $[a,b]$, the argument above can be localized to show that for some $A_i$, $\overline{A_i}\cap[a,b]$ contains an interval.

Answer 2

In response to the latest edit

Suppose that the Axiom of Choice holds. Then a countable union of countable sets is countable. As $ \mathbb{R} $ is uncountable, it follows that $ \mathbb{R} $ cannot be a countable union of countable sets.

There exists a model $ \mathcal{M} $ of the Zermelo-Fraenkel (ZF) axioms whose set of real numbers $ \mathbb{R}^{\mathcal{M}} $ is a countable union of countable sets. This is Theorem 10.6 of Thomas Jech’s The Axiom of Choice. Clearly, the Axiom of Choice must fail in $ \mathcal{M} $.

Therefore, you should be asking:

For all models $ \mathcal{M} $ of ZF, if $ \mathbb{R}^{\mathcal{M}} $ is a countable union of countable sets, is it true that one of those countable sets has to be dense?

Answer 3

The answer is surely no. Take a countable number $(\alpha_i)$ of irrational numbers in $[0,1]$ independent over $\mathbb Q$. Then let $X_i$ be the set of all elements of $[0,1]$ that equal $\alpha_i$ mod $\mathbb Q$. Then the $X_i$ are all countable. The union lies in $[0,1]$ so it cannot be dense in $\mathbb R$.

Note that, in my example, all the $X_i$ are disjoint (which you didn't require).

Answer 4

If you want the union of the countable sets to be dense in $\mathbb R$, take the union to be $\mathbb Q$. Enumerate the rationals in $[0,1)$ as $q_i$. Now let $S_i=q_i+z, z \in \mathbb Z$. Each $S_i$ looks like the integers, so is not dense, but the union is all of $\mathbb Q$.

Answer 5

To answer the question as modified by myself (a second time), even if $\bigcup A_i=\mathbf R$ and $i$ ranges over a countables set, there is no reason for any of the $A_i$ to be dense. Consider $A_i=\{\,x\in\mathbf R\mid \lfloor x\rfloor=i\,\}$ for $i\in\mathbf Z$.

Answer 6

This is an answer to the original question, asking whether a countable family of countable sets must have a member that is dense in $\mathbb{R}$. (I have answered the question under the restriction that their union is dense in $\mathbb{R}$, even though it was not in the original question.)

No. For any natural number $n$, take $X_n$ to be the set of rational numbers in $(n,n+1)$. Then the $X_n$ form a counterexample (indeed, the $X_n$ contain only positive numbers so are not dense in $\mathbb{R}$).

Answer 7

Try $A_i=[-i,i]$ $ $