Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 2 / 2G (Topics in Analysis)

A set is of first Baire category, or meagre, when it is a countable union of nowhere dense sets. The statement is true: a countable union of countable unions of nowhere dense sets is again a countable union of nowhere dense sets.

The relevant Baire category theorem says that a nonempty complete metric space is not meagre in itself.

False. Fix $y_0\in \mathbb{R}$ and takeThis closed set has empty interior in ${\mathbb{R}}^2$, so it is nowhere dense and hence meagre. Its section at $y_0$ is $C_{y_0}=\mathbb{R}$, which is not meagre in itself by the Baire category theorem.

True. Write $C={\bigcup }_{n\ge 1}{C}_n$ with every $C_n$ nowhere dense in $\mathbb{R}$. ThenThe closure of $C_n\times \mathbb{R}$ is
$\overline{C_n}\times \mathbb{R}$, which has empty interior because $\overline{C_n}$ has empty interior. Each product is therefore nowhere dense, so $A$ is meagre in ${\mathbb{R}}^2$.

False. The coordinate axesare both closed nowhere dense subsets of ${\mathbb{R}}^2$, but every $(x,y)$ is $(x,0)+(0,y)$, sowhich is not meagre by the Baire category theorem.