= Solution
For a fixed nonzero second coordinate $c$, the rational expression defining $h(x,c)$ is <continuous> on the half-line where $xc>0$. On the opposite half-line it is constant. At $x=0$ the ratio tends to $1$, agreeing with the value on the other side. If $c=0$ the entire restriction is constant. By symmetry the restrictions with the first coordinate fixed are <continuous> too. Hence $h$ is <separately continuous>, and so is $h_t(x,y)=h(x-t,y-t)$ for every real $t$.
On the diagonal $D$, $h_t(t,t)=1$, while $h_t(s,s)=0$ for $s\ne t$. Therefore
$$
D\cap h_t^{-1}((1/2,3/2))=\{(t,t)\}.
$$
Each diagonal <singleton> is open in the induced <topology>; arbitrary unions show that \b[$D$ has the <discrete topology>].
If $X$ were <metrizable space>[metrizable], its <countable> <dense subset> from the preceding part would make it <second countable>. Indeed balls with rational radii about <dense> points contain a <countable> <basis of a topology>[basis] after the usual smaller-ball argument. Every subspace of a <second-countable space> is <second countable>. A discrete <second-countable space> is <countable>, since each <singleton> must itself occur as a <basis of a topology>[basis] member. But $D$ is <uncountable>. This contradiction proves \b[$X$ is not <metrizable space>[metrizable]].
Back to article page
