Let $(Z,T_2)$ be a subspace of $R^2$ given by $Z = \{\langle x,y\rangle : x,y \in \Bbb R, y > 0\} \cup \{\langle x,0\rangle : x \in \Bbb Q\}$. Show that this is a Baire space. It is suggesting to use the fact that the closure of a Baire space is a Baire space.
I figure that the only dense subsets of $Z$ are in the upper half plane, but I can't figure out a way to write an arbitrary dense set. I figure $Z$ \ $A$, where $A$ is some collection of countable/uncountable plane curves in $R^2$ would work, but I want to see if there's a better way of writing a dense set.
The closure of a Baire space is Baire is short for the following theorem:
Suppose $D$ is dense in $(X,\mathcal{T})$ and suppose that $D$ is Baire in the subspace topology. Then $X$ is Baire.
Proof: let $U_n \subseteq X$ be open and dense. Then consider $V_n = D \cap U_n$ for all $n$. The $V_n$ are by definition open in $D$, they are non-empty as $D$ is dense and $V_n$ is non-empty open, and they are dense in $D$, because of the identity $\overline{U \cap D} = \overline{U}$, which is well-known for open and dense sets.
It follows that by assumption $\bigcap V_n$ is dense in $D$ and thus dense in $X$ ("dense in dense is dense"), and so the larger subset $\bigcap_n U_n$ is also dense in $X$, as required. QED.
Now note that the upper half plane is Baire (why?) and dense in your $Z$.
Also note that the closed subspace $\{(x,0): x \in \mathbb{Q} \}$ is not Baire, being homeomorphic to the rationals, which are not Baire.
So in contrast to open subspaces of Baire spaces (which are Baire again), closed subspaces need not be.