Define what it means for \((X,d)\) to be a metric space. Then \(d:X\times X\to\mathbb R\): is \(d\) continuous? Discuss. If you cannot answer in general do it for \(X=\mathbb R\) with the usual topology.
Solution
1. A metric on a set \(X\) is a function\[ d:X\times X\longrightarrow[0,\infty) \]such that for all \(x,y,z\in X\):\[ \begin{aligned} &d(x,y)=0\iff x=y,\\ &d(x,y)=d(y,x),\\ &d(x,z)\le d(x,y)+d(y,z). \end{aligned} \]A metric space is a pair \((X,d)\) equipped with the topology generated by its open balls.
These are respectively positive definiteness, symmetry, and the triangle inequality, together with the metric topology associated to \(d\).
2. For all \(x,x',y,y'\in X\),\[ \bigl|d(x,y)-d(x',y')\bigr| \le d(x,x')+d(y,y'). \]
The triangle inequality gives \[ d(x,y) \le d(x,x')+d(x',y')+d(y',y), \] so by symmetry, \[ d(x,y)-d(x',y') \le d(x,x')+d(y,y'). \] Interchanging \((x,y)\) with \((x',y')\) gives \[ d(x',y')-d(x,y) \le d(x,x')+d(y,y'). \] Combining these two inequalities yields the claimed absolute-value estimate.
3. The function\[ d:X\times X\longrightarrow\mathbb R \]is continuous for the product topology on \(X\times X\).
Fix \((x,y)\in X\times X\) and let \(\epsilon>0\). Set \[ U=B(x,\epsilon/2), \qquad V=B(y,\epsilon/2). \] Then \(U\times V\) is an open neighborhood of \((x,y)\) in the product topology. For every \((x',y')\in U\times V\), 2 gives \[ \begin{aligned} \bigl|d(x',y')-d(x,y)\bigr| &\le d(x',x)+d(y',y)\\ &<\frac\epsilon2+\frac\epsilon2\\ &=\epsilon. \end{aligned} \] Thus \[ d(U\times V)\subseteq (d(x,y)-\epsilon,d(x,y)+\epsilon). \] This is continuity of \(d\) at \((x,y)\), and the point was arbitrary.
4. Hence the answer is yes for every metric space, not merely for \(X=\mathbb R\).
This is exactly 3.