Past exam of the options I course of the University of Oxford / 2012 / E / E1

The map is $K$-Lipschitz continuous whenThe same finite Lipschitz constant $K$ must work for all pairs of points.

The map is uniformly continuous when, for every $ϵ>0$, there is a $\delta >0$ such thatThe choice of $\delta$ may depend on $ϵ$ but must be independent of the locations of $x$ and $x^{\prime }$.

Given $ϵ>0$, choose $\delta =ϵ/K>0$. If $d_X(x,x^{\prime })<\delta$, the Lipschitz condition yieldsThis single $\delta$ works for every pair, proving the map is uniformly continuous:

The distance to a set is finite and nonnegative because $A$ is nonempty. For $x,x^{\prime }\in X$ and every $a\in A$, the triangle inequality givesTaking the infimum over $a$ yields $D_A(x)\le d(x,x^{\prime })+D_A(x^{\prime })$. Interchanging $x$ and $x^{\prime }$ gives the opposite inequality. Together these implyThus the distance to a set is $1$-Lipschitz continuous. No closedness, compactness or attainment of the infimum was used.

For fixed $x\in X$, the function $a\mapsto d(x,a)$ on $A$ is continuous, indeed $1$-Lipschitz continuous by the reverse triangle inequality. A continuous real-valued function on a nonempty compact space attains its minimum by the extreme value theorem. Hence some $a_0\in A$ satisfiesThe metric is symmetric, so this is also $d(a_0,x)$. Compactness supplies attainment of the infimum.