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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?

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Does countable imply infinite for you? Have you tried some examples? –  Daan Michiels Feb 4 '13 at 15:26
Uncountable sets can be nowhere dense (like the Cantor set), so we can partition the Cantor Set into an uncountable number of disjoint countable sets using the Axiom of Choice--none of them can be dense. –  Andrew Salmon Feb 4 '13 at 15:29
Presumably you want the union to be dense in $\mathbb R$ –  Ross Millikan Feb 4 '13 at 15:36
Yes, it seems something is missing... –  Daan Michiels Feb 4 '13 at 15:37
Sorry! What I ment is: Given a countable union of sets in R, $A_i$ such that $\cup A_i = R$, Must one of them be dense in R? of course, if the union is not R one can look at the integers as a countable union of countable sets. sorry for not pointing that out in the first time. Or another question that mught help me: if a set is not countable, must it be dense in some open segment of R? Thanks! –  Shir Sivroni Feb 5 '13 at 11:36
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7 Answers

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.

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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?

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This answer makes it sound that the two things are equivalent, while they are clearly not. The axiom of choice may fail acutely and the real numbers are not a countable union of countable sets. –  Asaf Karagila Feb 5 '13 at 12:56
@Asaf: I’ve clarified my presentation in response to your comment. Thanks! –  Haskell Curry Feb 5 '13 at 16:26
Now it's much more accurate! :-) I still think that this answer shoots way way over the OP's head. –  Asaf Karagila Feb 5 '13 at 16:31
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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).

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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 $\mathbb Q$.

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Your construction also gives disjoint sets $S_i$, very nice. –  hardmath Feb 4 '13 at 16:03
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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 $i\in\mathbf Z$.

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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 $\mathbb{R}$).

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These are each dense somewhere, but you are correct they are not dense in all of $mathbb R$ –  Ross Millikan Feb 4 '13 at 15:36
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Try $A_i=[-i,i]$ $ $

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