Past exam of the options I course of the University of Oxford / 2004 / E

A metric space is a set $X$ with a function $d:X\times X\to [0,\infty )$ satisfying $d(x,y)=0⟺x=y$, symmetry and the triangle inequality. A subset $U$ is open if every $x\in U$ has some open open ball $B(x,r)$ contained in $U$.

For finitely many open sets $U_1,\dots ,U_k$ and $x$ in their intersection, choose radii $r_j>0$ with $B(x,r_j)\subseteq U_j$. The positive minimum radius gives a ball contained in the intersection. Thus every finite intersection is open; the empty intersection is $X$, also open.

For distinct points put $\delta =d(x_1,x_2)>0$. The balls $B(x_1,\delta /3)$ and $B(x_2,\delta /3)$ are disjoint: a common point would make $d(x_1,x_2)<2\delta /3$ by the triangle inequality. This proves the Hausdorff condition.

A singleton $\{x\}$ is closed because each $y\ne x$ has a ball of radius $d(x,y)/2$ avoiding it, making its complement open. Finite unions of closed sets are closed, as their complements are finite intersections of open sets. Hence every finite subset is closed, including the empty subset.

The product topology is generated by the basis of a topology of open rectangles $U\times V$, with $U$ open in $X$ and $V$ open in $Y$. For a fixed $y$,Both possibilities are open; checking this basis of a topology proves that $i_y$ is continuous. The analogous map $j_x:Y\to X\times Y$, $j_x(y)=(x,y)$, is continuous by the same argument.

If either factor is empty the product is empty and connected. Otherwise let $h:X\times Y\to \{0,1\}$ be continuous. For each $y$, $h\circ i_y$ is constant since $X$ is connected. Thus $h(x,y)=h(x_0,y)$ for any fixed $x_0\in X$. The right-hand side is $h\circ j_{x_0}$ and is constant in $y$ since $Y$ is connected. Therefore $h$ is constant on the whole product. The given characterization proves $X\times Y$ is connected.