# Is the complement of a countable set in $\mathbb{R}$ dense? Application to convergence of probability distribution functions.

I am wondering if we have a set $A\in\mathbb{R}$ that is countable, whether $A^{c}$ is dense in $\mathbb{R}$? I thought I saw this quoted somewhere on google but I can not find it again! I am working through a proof relating to uniqueness of weak limits of sequences distribution functions. The set of discontinuity points of distribution functions are countable, and the proof suggests the complement of this set is dense in $\mathbb{R}$, and I am unsure how this conclusion is arrived at.

Any help would be greatly appreciated.

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Yes it is. Suppose it's not : then there exists an nonempty open set $U$ of $\mathbb{R}$ which doesn't meet $A^C$. So $(A^C)^C = A$ contains a non empty open interval, and since open interval are in bijection with $\mathbb{R}$, $A$ is not countable.
Thanks for the immediate reply, and thanks for the nice clear proof of why $A^{c}$ is indeed dense. Presumably by two sets not meeting you mean they are disjoint? – dandar Mar 12 '12 at 21:43
I think you did use the correct term. I think that 2 sets meeting is a mathematical term that has something to do with lattices, but perhaps for sets in $\mathbb{R}$ means they are not disjoint. The book I am reading uses the same term but does not describe what it means. You have confirmed at least that you also thought as I did. Thanks again. – dandar Mar 13 '12 at 18:12
Let $a\in \Bbb R$ and suppose $U$ is an open set containing $a$. Then since $U$ contains uncountably many points in $\Bbb R$ and since $A$ is countable, $U\setminus A$ is uncountable and thus contains points of $A^C$ distinct from $a$. It follows that $a$ is in the closure of $A^C$. As $a$ was an arbitrary point in $\Bbb R$, it follows that $A^C$ is dense in $\Bbb R$.