Irrational number and Baire space How to show that the set of irrational numbers is a Baire space ?
 A: The Baire category theorem gives sufficient conditions for a topological space to be a Baire space. It is an important tool in topology and functional analysis.
    (BCT1) Every complete metric space is a Baire space. More generally, every topological space which is homeomorphic to an open subset of a complete pseudometric space is a Baire space. In particular, every completely metrizable space is a Baire space.
    (BCT2) Every locally compact Hausdorff space is a Baire space.
    BCT1 shows that each of the following is a Baire space:
    The space R of real numbers;
    The space of irrational numbers;
    The Cantor set;
    Indeed, every Polish space;
http://en.wikipedia.org/wiki/Baire_space
http://en.wikipedia.org/wiki/Complete_metric_space
another definition :
A topological space X in which each subset of X of the "first category" has an empty interior. A topological space which is homeomorphic to a complete metric space is a Baire space, ,where terminology homeomorphic :
two objects are homeomorphic if they can be deformed into each other by a continuous, invertible mapping
A: Let's see. The definition of a Baire space is that a countable intersection of dense open sets is dense. So let $I$ be the space of irrational numbers, and let $U_1,U_2,U_3,\dots$ be a sequence of dense open subsets of $I$; I have to show that $\bigcap_{n=1}^\infty U_n$ is dense. Now, "$U$ is open in $I$" means that $U=V\cap I$ for some $V$ which is open in $\mathbb R$; moreover, if $V\cap I$ is dense in $I$, then (since $I$ is dense in $\mathbb R$) it's dense in $\mathbb R$, and so is $V$. So it's enough to show is that, if $V_1,V_2,\dots$ is a sequence of dense open sets in $\mathbb R$, then $\bigcap_{n=1}^\infty(V_n\cap I)=(\bigcap_{n=1}^\infty V_n)\cap I$ is dense. But $I$ itself is a countable intersection of dense open subsets of $\mathbb R$, namely, $I=\bigcap_{q\in\mathbb Q}(\mathbb R\setminus\{q\})$. So now I have to show that $(\bigcap_{n=1}^\infty V_n)\cap\bigcap_{q\in\mathbb Q}(\mathbb R\setminus\{q\})$ is dense, but that's a countable intersection of dense open subsets of $\mathbb R$, so by the Baire category theorem . . .
