Distance to a set ($d(x,F)={\inf }_{y\in F}d(x,y)$)

For a nonempty subset $F$ of a metric space, the distance to a set is $d(x,F)={\inf }_{y\in F}d(x,y)$. The triangle inequality gives $d(x,F)\le d(x,z)+d(z,F)$; swapping $x,z$ proves $|d(x,F)-d(z,F)|\le d(x,z)$. Thus it is one-Lipschitz. Its zero set is exactly the closure of $F$: every neighbourhood of $x$ meets $F$ precisely when this infimum vanishes. For a closed set, zero distance therefore implies membership. When the empty set is allowed, use the convention $d(x,\varnothing )=+\infty$; the finite-valued Lipschitz assertion applies to nonempty sets.

The set distance is the infimum of pairwise metric distances. Two nonempty disjoint compact sets have positive set distance: continuity attains the minimum on their Cartesian product, and a zero minimum would give a common point. Disjoint closed sets without compactness need not have positive separation.

In a metric space $X$, let the nonempty compact set $K$ lie in the open set $\Omega$. If $F=X\setminus \Omega$ is nonempty, the distance to a set $d(x,F)$ is a continuous function of $x$: the triangle inequality gives $|d(x,F)-d(y,F)|\le d(x,y)$. Since $F$ is a closed set, this distance is strictly positive on $K$. The extreme value theorem attains its minimum on $K$, soIf $F$ is empty, the extended-real convention gives $d(K,F)=+\infty$. This result provides a single positive neighbourhood radius that stays inside $\Omega$ at every point of $K$. Compactness is essential: in the real line $K=\{1/n:n\ge 1\}\subset (0,2)$ has zero distance between two sets from $(-\infty ,0]\cup [2,\infty )$, although each point of $K$ has positive distance to a set from that complement.