Let \(X\) be a metric space with metric \(d\).
Show that \(d: X \times X \to \mathbb{R}\) is continuous.
Let \(X'\) denote a space having the same underlying set as \(X\). Show that if \(d: X' \times X' \to \mathbb{R}\) is continuous, then the topology of \(X'\) is finer than the topology of \(X\).
One can summarize the result of this exercise as follows: if \(X\) has a metric \(d\), then the topology induced by \(d\) is the coarsest topology relative to which the function \(d\) is continuous.
Solution
The triangle inequality gives the reverse-triangle estimate \[ |d(x,y)-d(x',y')| \le d(x,x')+d(y,y'). \] Indeed \[ d(x,y)\le d(x,x')+d(x',y')+d(y',y), \] and reversing the two pairs gives the opposite inequality. Therefore, given \(\varepsilon>0\), if \[ d(x,x')<\varepsilon/2,\qquad d(y,y')<\varepsilon/2, \] then \[ |d(x,y)-d(x',y')|<\varepsilon. \] Thus \(d:X\times X\to\mathbb R\) is continuous.
Assume \(d:X'\times X'\to\mathbb R\) is continuous, where \(X'\) has the same underlying set. Fix \(a\in X'\). The map \[ j_a:X'\to X'\times X',\qquad j_a(x)=(a,x) \] is continuous, so \[ x\longmapsto d(a,x)=d\circ j_a(x) \] is continuous. Hence for every \(r>0\) the metric ball \[ B_d(a,r)=\{x:d(a,x)<r\} \] is open in \(X'\). Since metric balls form a basis for the topology of \(X\), every \(X\)-open set is \(X'\)-open. Therefore the topology of \(X'\) is finer than the metric topology of \(X\).