Questions on meager sets

Question

I want to know if I answered them correctly:

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

1. 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.

2. 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.

3. 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.

Answer 1

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.