= Solution
Fix a horizontal line $L=\mathbb R\times\{c\}$. The restriction of every member of $\mathscr H$ to $L$ is <continuous> in its usual <topology>. Therefore every subbasic <open set> intersects $L$ in a usual <open set>, so the induced <topology> is no finer than the usual one. Conversely the first <projection map>[coordinate projection] is <continuous> for the <initial topology>, and its inverse images induce every usual <open set> on $L$. The two <topologies> on $L$ are equal. The second projection gives the identical argument for a vertical line.
Write $F=\overline Y^{\,X}$. For every nonzero rational $r$, the <intersection> of $Y$ with $\{r\}\times\mathbb R$ is <dense> in that vertical line, since the nonzero <rational numbers> are <dense> in the <real line>. Hence $\{r\}\times\mathbb R\subseteq F$. For each real $c$, it follows that $F\cap(\mathbb R\times\{c\})$ contains $(\mathbb Q\setminus\{0\})\times\{c\}$ and is closed in that line. This <dense subset> forces the whole horizontal line into $F$. Thus $F=X$: \b[$Y$ is a <countable> <dense subset>]. The two successive uses of line density are necessary; the <initial topology> is finer than the <Euclidean topology>, so Euclidean density alone would not suffice.
Back to article page
