Past exam of the analytic topology and axiomatic set theory course of the University of Oxford / 2009 / A / 1 / iii

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 $s\ne t$. ThereforeEach diagonal singleton is open in the induced topology; arbitrary unions show that $D$ has the discrete topology.

If $X$ were 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 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 member. But $D$ is uncountable. This contradiction proves $X$ is not metrizable.