 Compact subset of an open set has positive distance from its complement

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.