Let \((X,d)\) be a metric space and let \(A\subset X\). If \(x\in X\) define the distance of \(x\) to \(A\) to be \(\inf\{d(x,a):a\in A\}\). Prove that the real-valued function on \(X\) defined by \(x\mapsto d(x,A)\) is continuous.
Solution
The statement is meaningful as a real-valued function when \(A\ne\varnothing\), which we assume below.
1. For all \(x,y\in X\),\[ d(x,A)\le d(x,y)+d(y,A). \]
Fix \(x,y\in X\). For every \(a\in A\), the triangle inequality gives \[ d(x,a)\le d(x,y)+d(y,a). \] Since \[ d(x,A)=\inf_{b\in A}d(x,b)\le d(x,a), \] we have \[ d(x,A)\le d(x,y)+d(y,a) \] for every \(a\in A\). Thus \(d(x,A)\) is a lower bound for the nonempty set \[ \{d(x,y)+d(y,a):a\in A\}. \] It is therefore at most its infimum, namely \[ d(x,y)+\inf_{a\in A}d(y,a) =d(x,y)+d(y,A). \]
2. The distance-to-\(A\) function is \(1\)-Lipschitz:\[ |d(x,A)-d(y,A)|\le d(x,y) \]for all \(x,y\in X\).
From 1, \[ d(x,A)-d(y,A)\le d(x,y). \] Interchanging \(x\) and \(y\) and using symmetry of the metric gives \[ d(y,A)-d(x,A)\le d(y,x)=d(x,y). \] Together these inequalities are equivalent to \[ |d(x,A)-d(y,A)|\le d(x,y). \]
3. The function\[ f:X\to\mathbb R, \qquad f(x)=d(x,A), \]is continuous.
Let \(x\in X\) and let \(\varepsilon>0\). Take \[ \delta=\varepsilon. \] If \(d(x,y)<\delta\), then by 2, \[ |f(x)-f(y)| =|d(x,A)-d(y,A)| \le d(x,y) <\delta =\varepsilon. \] Thus \(f\) is continuous at every \(x\in X\); indeed, the same estimate shows that it is uniformly continuous.