Questions on meager sets I want to know if I answered them correctly:

Let $(X,d)$ be an arbitrary complete metric space. Prove or disprove

*

*For any $x\in X$ the subset  $\{x\}\subset X$ is meager


Wrong. Let $X=\mathbb{Z}$ with the discrete metric. Then $\{z\}\in \mathbb{Z}$ is not meager.



*If $U$ is nonempty open subset of $X$, then $U$ is not meager


Wrong. $U=\mathbb{Q}$ is an nonempty open subset of $X=\mathbb{R}$ which is meager.



*Any subset of $X$, which is dense in $X$ is residual


True then a residual set is the complement of a meager set and the complement of a meager set can not be nowhere dense.
 A: The first one is correct; the other two are not. 
For (2) suppose that $U$ is meagre. Then there are nowhere dense sets $A_n$ for $n\in\Bbb N$ such that $U=\bigcup_{n\in\Bbb N}A_n$, so $U\subseteq\bigcup_{n\in\Bbb N}\operatorname{cl}A_n$. For $n\in\Bbb N$ let $V_n=X\setminus\operatorname{cl}A_n$; $\operatorname{cl}A_n$ is a closed nowhere dense set in $X$, so $V_n$ is a dense open subset of $X$. $X$ is complete, so $\bigcap_{n\in\Bbb N}V_n$ is dense in $X$. But then on the one hand $U\cap\bigcap_{n\in\Bbb N}V_n\ne\varnothing$, since $U$ is a non-empty open set, and on the other hand
$$U\cap\bigcap_{n\in\Bbb N}V_n=U\cap\bigcap_{n\in\Bbb N}(X\setminus\operatorname{cl}A_n)=U\cap\left(X\setminus\bigcup_{n\in\Bbb N}A_n\right)=U\setminus\bigcup_{n\in\Bbb N}A_n=\varnothing\;.$$
This contradiction shows that $U$ cannot be meagre. The specific problem with your example is that $\Bbb Q$ is not an open set in $\Bbb R$.
For (3) let $X=\Bbb R$ with the usual metric; $\Bbb R$ is complete, and $\Bbb Q$ is dense in $\Bbb R$. If $\Bbb Q$ were residual in $\Bbb R$, $\Bbb R\setminus\Bbb Q$ would be meagre, and there would be nowhere dense sets $A_n$ in $X$ such that 
$$\Bbb R\setminus\Bbb Q=\bigcup_{n\in\Bbb N}A_n\;.$$
But then
$$\Bbb R=\bigcup_{q\in\Bbb Q}\{q\}\cup\bigcup_{n\in\Bbb N}A_n$$
would be the union of countably many nowhere dense sets and would be meagre in itself. Show that this is impossible by using the Baire category theorem to show that a complete metric space is never meagre in itself.
