Past exam of the mathematics course of the University of Cambridge / 2010 / ii / Paper 1 / 2F (Topics in Analysis)

The Baire category theorem says that a countable intersection of open dense sets in a nonempty complete metric space is dense. Write $E=\{e_1,e_2,\dots \}$. Each singleton is closed in a metric space and has empty interior because there are no isolated points. Therefore every $X\setminus \{e_j\}$ is open and dense. Applying the theorem also to $G$ yieldswhich is dense in $X$. Thus every nonempty open subset meets the required difference, not merely the original dense set $G$.

No. Take the complete metric space $X=\mathbb{R}$ with its usual metric, $E=\mathbb{Q}$, and $G=\mathbb{Q}\cup (1,2)$. This $G$ is uncountable and dense, whereasmisses, for example, the nonempty open interval $(3,4)$. Uncountability does not replace the open-dense hypothesis in the Baire category theorem.