Countable and dense Baire subsets I'm having trouble with an excercise, which says:
Let $A\subset \mathbb{R} $ be a Baire space. Then $A$ can't be countable and dense simultaneously.
I think I have found a proof, but it doesn't involve using the fact that $A$ is dense, just that it's countable and that it's a subset of $\mathbb{R}$. Is there a counterxample to that? That is, a countable Baire space in $\mathbb{R}$ that isn't dense? 
 A: The main idea: if $A$ is dense in $\mathbb{R}$ it has no isolated points (if $x \in A$ were isolated there would be an open interval $(c,d) \subseteq \mathbb{R}$ with $(c,d) \cap A = \{x\}$, but then $(c,x)$ is a non-empty open set of the reals that is disjoint from $A$, so $A$ is not dense in the reals).
If then $A$ were also countable, we can write it as a countable union of singletons $\{a\}$, each of which is then closed (all singetons are closed in metric spaces) and nowhere dense in $A$ (as no point is isolated in $A$) So $A$ cannot be Baire (all sets $A \setminus \{a\}$ are open and dense in $A$ and their intersection is empty, so not dense). 
So if we assume that $A$ is Baire it cannot be both dense and countable, which is what was to be shown.
As to the final question: $\mathbb{Z}$ is Baire and countable (but not dense, which makes sense as it only consists of isolated points). Also $\{0\} \cup \{\frac{1}{n}: n=1,2,3\ldots\}$ is such a space, with only one non-isolated point $0$, to see that non-discrete examples exist too.
